Algebraic K-Theory and its Applications [1 (first printing) ed.] 9780387942483, 0387942483, 9783540942481, 3540942483

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Graduate Texts in Mathematics

147

Editorial Board J.H. Ewing F.W. Gehring P.R. Halmos

Graduate Texts in Mathematics 1 TAKEUTIEARING. Introduction to Axiomatic Set Theory. 2nd ed. 2 OXTOBY. Measure and Category. 2nd ed. 3 SCHAEFFER. Topological Vector Spaces. 4 HILTON/~TAMMBACH. A Course in Homological Algebra. 5 MAC LANE. Categories for the Working Mathematician. 6 HUGHES/PIPER. Projective Planes. 7 SERRE. A Course in Arithmetic. 8 TAKEUTI/ZAR~NG. Axiometic Set Theory. 9 HUMPHREYS. Introduction to Lie Algebras and Representation Theory. 10 COHEN. A Course in Simple Homotopy Theory. 11 CONWAY. Functions of One Complex Variable. 2nd ed. 12 BEALS. Advanced Mathematical Analysis. 13 ANDERSON /FULLER. Rings and Categories of Modules. 2nd ed. 14 GOLUBITSKY/GUILE.MIN. Stable Mappings and Their Singularities. 15 BERBERIAN. Lectures in Functional Analysis and Operator Theory. 16 WINTER . The Structure of Fields. 17 ROSENBLA-~T. Random Processes. 2nd ed. 18 HALMOS. Measure Theory. 19 HALMOS . A Hilbert Space Problem Book. 2nd ed. 20 HUSEMOLLER. Fibre Bundles. 3rd ed. 21 HUMPHREYS. Linear Algebraic Groups. 22 BARNESMACK. An Algebraic Introduction to Mathematical Logic. 23 GREUB . Linear Algebra. 4th ed. 24 HOLMES . Geometric Functional Analysis and Its Applications. 25 HEWI-~TETROMBERG. Real and Abstract Analysis. 26 MANES . Algebraic Theories. 27 KELLEY . General Topology. 28 ZARISKIISAMUEL. Commutative Algebra. Vol. I. 29 ZARISK~EAMUEL. Commutative Algebra. Vol. II. 30 JACOBSON . Lectures in Abstract Algebra I. Basic Concepts. 31 JACOBSON . Lectures in Abstract Algebra II. Linear Algebra. 32 JACOBSON . Lectures in Abstract Algebra III. Theory of Fields and Galois Theory. 33 HIRSCH . Differential Topology. 34 SPITZER. Principles of Random Walk. 2nd ed. 35 WERMER . Banach Algebras and Several Complex Variables. 2nd ed. 36 KELLEYINAMIOKA et al. Linear Topological Spaces. 37 MONK. Mathematical Logic. 38 GRAUERTERIEXHE. Several Complex Variables. 39 ARVESON. An Invitation to C*-Algebras. 40 KEMENY/SNELLENAPP. Denumerable Markov Chains. 2nd ed. 41 APOSTOL . Modular Funct~~~~~,~~~!~~_~e~~~_in,N_umber Theory. 2nd ed. 42 SERRE. Linear Represenmti ti!@ep,ups.“-$nz *u&a-Fut@ions 43 GILLMAN/JERISON. Rings) of@?Y 44 KENDIG. Elementary Al&b& Geomett$ 45 LO~VE. Probability The% Ii&h ed>( 46 Lo&z. Probability Theoh II:%th ed: 47 MOISE. Geometric Topold~~~~~~im~~~~~s~~ ‘a&J 3.

Jonathan Rosenberg

Algebraic K-Theory and Its Applications

Springer-Verlag New York Berlin Heidelberg London Paris Tokyo Hong Kong Barcelona, Budapest,

Jonathan Rosenberg Department of Mathematics University of Maryland College Park, MD 20742 USA Editorial Board J.H. Ewing Department of Mathematics Indiana University Bloomington, IN 47405 USA

F.W. Gehring Department of Mathematics University of Michigan Ann Arbor, MI 48109 USA

P.R. Halmos Department of Mathematics Santa Clara University Santa Clara, CA 95053 USA

Mathematics Subject Classifications (1991): 19.01, 19-02, 19AXX, 19BXX, 19CXX, 55QXX, lSEXX, 19FXX, 19JXX, 19KXX, 18G60

Library of Congress Cataloging in Publication Data Rosenberg, J. (Jonathan), 1951Algebraic K-theory and its applications / Jonathan Rosenberg. p. cm. -- (Graduate texts in mathematics ; ) Includes bibliographical references and index. ISBN O-387-94248-3 1. K-theory. I. Title. II.Series. QA612.33.R67 1994 94-8077 512 .55--dc20

Printed on acid-free paper. 0 1994 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone. Production managed by Karen Phillips; manufacturing supervised by Gail Simon. Photocomposed pages prepared from the author’s AMSTEX file. Printed and bound by R.R. Donnelley and Sons, Harrisonburg, VA. Printed in the United States of America. 987654321

ISBN O-387-94248-3 Springer-Verlag New York Berlin Heidelberg ISBN 3-540-94248-3 Springer-Verlag Berlin Heidelberg New York

Preface

Algebraic K-theory is the branch of algebra dealing with linear algebra (especially in the limiting case of large matrices) over a general ring R instead of over a field. It associates to any ring R a sequence of abelian groups Ki(R). The first two of these, Ko and K1, are easy to describe in concrete terms; the others are rather mysterious. For instance, a finitely generated projective R-module defines an element of Ko( R), and an invertible matrix over R has a “determinant” in Kl(R). The entire sequence of groups Ki(R) behaves something like a homology theory for rings. Algebraic K-theory plays an important role in many areas, especially number theory, algebraic topology, and algebraic geometry. For instance, the class group of a number field is essentially Ko(R), where R is the ring of integers, and “Whitehead torsion” in topology is essentially an element of KI(ZT), where 7r is the fundamental group of the space being studied. K-theory in algebraic geometry is basic to Grothendieck’s approach to the Riemann-Roth problem. Some formulas in operator theory, involving determinants and determinant pairings, are best understood in terms of algebraic K-theory. There is also substantial evidence that the higher K-groups of fields and of rings of integers are related to special values of L-functions and encode deep arithmetic information. This book is based on a one-semester course I gave at the University of Maryland in the fall of 1990. Most of those attending were second- or third-year graduate students interested in algebra or topology, though there were also a number of analysis students and faculty colleagues from other areas. I tried to make the course (and this book) fairly self-contained, and to assume as a prerequisite only the standard one-year graduate algebra course, based on a text such as [Hungerford], [Jacobson], or [Lang], and the standard introductory graduate course on algebraic and geometric topology, covering the fundamental group, homology, the notions of simplicial and CW-complex, and the definition and basic properties of manifolds. As taught at Maryland, the graduate algebra course includes the most basic definitions and concepts of category theory; a student who hasn’t yet seen these ideas could consult any of the above algebra texts or an introduction to category theory such as [Mac Lane]. Since many graduate algebra courses do not include much in the way of algebraic number theory, I have

vi

Preface

included many topics such as the basic theory of Dedekind rings and the Dirichlet unit theorem,. which may be familiar to some readers but not to all. I’ve tried in this book to presuppose as little topology as possible beyond a typical introductory course, and to develop what is needed as I go along, but to give the reader a flavor of some of the important applications of the subject. A reader with almost no topology background should still be able to follow most of the book except for parts of Sections 1.6, 1.7, 2.4, 4.4, and 6.3, and most of Chapter 5 (though I would hope this book might encourage him or her to take a more systematic course in topology). A problem one always has in writing a book such as this is to decide what to do about spectral sequences. They are usually not mentioned in first-year graduate courses, and yet they are indispensable for serious work in homological algebra and K-theory. To avoid having to give an introduction to spectral sequences which might scare off many readers, I have avoided using spectral sequences directly anywhere in the text. On the other hand, I have made indirect reference to them in many places, so that the reader who has heard of them will often see why they are relevant to the subject and how they could be used to simplify some of the proofs. For the most part, this book tends to follow the notes of the original course, with a few additions here and there. The major exceptions are that Chapters 3 and 5 have been greatly expanded, and Chapter 6 on cyclic homology has been added even though there was no time for it in the original course. Cyclic homology is a homology theory for rings which may be viewed as the “linearized version” of algebraic K-theory, and it’s becoming increasingly clear that it is both a useful computational tool and a subject of independent interest with its own applications. Each chapter of this book is divided into sections, and I have used a single numbering system for all theorems, lemmas, exercises, definitions, and formulas, to make them easier to locate. Thus a reference such as 1.4.6 means the 6th numbered item in Section 4 of Chapter 1, whether that item is a theorem, a corollary, an exercise, or a displayed formula. The exercises are an integral part of the book, and I have tried to put at least one interesting exercise at the end of every section. The reader should not be discouraged if he finds some of the exercises too difficult, since the exercises vary from the routine to the very challenging. I have used a number of more-or-less standard notations without special reference, but the reader who is puzzled by them will be able to find most of them listed in the Notational Index in the back of the book. Why This Book? The reader might logically ask how this book differs from its “competition.” [Bass] remains an important reference, but it is too comprehensive to use as a text for an elementary course, and also it predates the definition of K2, let alone of higher K-theory or of cyclic homology. My original course was based on the notes by Milnor [Milnor], which are highly recommended. However, I found that [Milnor] is hard to use as a textbook, for

Preface

vii

the following three reasons: (1) Milnor writes for a working mathematician, and sometimes leaves out details that graduate students might not be able to provide for themselves. (2) There are no exercises, at least in the formal sense. (3) The subject has changed quite a bit since Milnor’s book was written. For the working algebraist already familiar with the contents of [Milnor] who wants to learn about Quillen K-theory and its applications in algebraic geometry, [Srinivas] is an excellent text, but it would have been far beyond the reach of my audience. The notes of Berrick [Berrick] give a more elementary introduction to Quillen K-theory than [Srinivas], but are rather sketchy and do not say much about applications, and thus again are not too suitable for a graduate text. [LluisP] is very good for an up-to-date survey, but it is, as the title says, an overview rather than a textbook. For cyclic homology, the recent book by Loday [LodayCH] is excellent, but to be most useful it requires that the reader already know something about K-theory. Also, I do not believe that there is any book available that discusses the applications of algebraic K-theory in functional analysis (which are discussed here in 2.2.1G2.2.11, 4.4.19-4.4.24, 4.4.30, 6.3.8-6.3.17, and 6.3.296.3.30). Thus for all these reasons it seemed to me that another book on K-theory is needed. I hope this book helps at least in part to fulfill that need. Acknowledgments I would like to thank Mike Boyle for making his notes of my lectures (often much more readable and complete than my own) available to me, and to thank all the others who attended my lectures for their useful feedback. I would also like to thank the National Science Foundation for its support under grants # DMS 90-02642 and # DMS 92-25063, which contributed substantially to the research that went into the writing of this book, both directly and indirectly. Finally, I would like to thank several anonymous referees and numerous colleagues, including in particular Andrew Ranicki and Shmuel Weinberger, for useful suggestions on how to improve my early drafts.

Contents

Preface Chapter 1. K0 of Rings 1. Defining Ko 2. KO from idempotents 3. .K,-, of PIDs and local rings 4. Ko of Dedekind domains 5. Relative K~-J and excision 6. An application: Swan’s Theorem and topological K- theory 7. Another application: Euler characteristics and the Wall finiteness obstruction Chapter 2. K1 of Rings 1. Defining K1 2. Kl of division rings and local rings 3. KI of PIDs and Dedekind domains 4. Whitehead groups and Whitehead torsion 5. Relative K1 and the exact sequence Chapter 3. KO and K1 of Categories, Negative K-Theory 1. Ko and Kl of categories, GO and G1 of rings 2. The Grothendieck and Bass-Heller-Swan Theorems 3. Negative K-theory Chapter 4. Milnor’s Kz 1. Universal central extensions and HZ Universal central extensions Homology of groups 2. The Steinberg group 3. Milnor’s K2 4. Applications of KZ Computing certain relative KI groups K2 of fields and number theory Almost commuting operators Pseudo-isotopy

V

1 1 7 11 16 27 32 41 59 59 62 74 83 92 108 108 132 153 162 162 163 168 187 199 218 218 221 237 240

x

Contents

Chapter 5. The +-Construction and Quillen K-Theory 1. An introduction to classifying spaces 2. Quillen’s +-construction and its basic properties 3. A survey of higher K-theory Products K-theory of fields and of rings of integers The Q-construction and results proved with it Applications

245 245 265 279 279 281 289 295

Chapter 6. Cyclic homology and its relation to K-Theory 1. Basics of cyclic homology Hochschild homology Cyclic homology Connections with “non-commutative de Rham theory” 2. The Chern character The classical Chern character The Chern character on Ko The Chern character on higher K-theory 3. Some applications Non-vanishing of class groups and Whitehead groups Idempotents in C*-algebras Group rings and assembly maps

302 302 302 306 325 331 332 335 340 350 350 355 362

References

369

Books and Monographs on Related Areas of Algebra, Analysis, Number Theory, and Topology Books and Monographs on Algebraic K-Theory Specialized References

369 371 372

Notational Index

377

Subject Index

303

1 KO of Rings

1. Defining KO K-theory as an independent discipline is a fairly new subject, only about 35 years old. (See [Bak] for a brief history, including an explanation of the choice of the letter K to stand for the German word Klasse.) However, special cases of K-groups occur in almost all areas of mathematics, and particular examples of what we now call Ko were among the earliest studied examples of abelian groups. More sophisticated examples of the idea of the definition of Ko underlie the Euler-Poincar6 characteristic in topology and the Riemann-Roth theorem in algebraic geometry. (The latter, which motivated Grothendieck’s first work on K-theory, will be briefly described below in 53.1.) The Euler characteristic of a space X is the alternating sum of the Betti numbers; in other words, the alternating sum of the dimensions of certain vector spaces or free R-modules I&(X; R) (the homology groups with coefficients in a ring R). Similarly, when expressed in modern language, the Riemann-Roth theorem gives a formula for the difference of the dimensions of two vector spaces (cohomology spaces) attached to an algebraic line bundle over a non-singular projective curve. Thus both involve a formal difference of two free modules (over a ring R which can be taken to be Cc). The group Ko(R) makes it possible to define a similar formal difference of two finitely generated projective modules over any ring R. We begin by recalling the definition and a few basic properties of projective modules. Unless we say otherwise, we shall assume all rings have a unit, we shall require all ring homomorphisms to be unitpreserving, and we shall always use the word module to mean “l eft module . ” 1.1.1. Definition. Let R be a ring. A projective module over R means an R-module P with the property that any surjective R-module homomorphism Q : M + P has a right inverse p : P -+ M. An equivalent way of phrasing this is that whenever one has a diagram of R-modules and

2

1. KO of Rings

R-module maps

P

1 M

-

*

‘p

N

with M 2 N surjective, one can fill this in to a commutative diagram

* M N. Indeed, given the diagram-completion property and a surjective R-module homomorphism (Y : M ---f P, one can take N = P, cp = idp, and + = cq and the resulting 8 : P ---t M is a right inverse for (Y, i.e., satisfies a:08=idp. In the other direction, suppose any surjective R-module homomorphism CX: M + P has a right inverse p : P + M, and suppose one is given a diagram of R-modules and R-module maps P

with M -% N surjective. Replacing M 2 N by M @ P s N 63 P and cp : P -+ N by (cp, idp) : P --+ N @ P, we may suppose ‘p is one-to-one, and then replacing N by the image of cp and M by $-‘(imcp), we may assume it’s an isomorphism. Then take (Y = ‘p-l o $ and the right inverse p : P + M enables us to complete the diagram. When cx : M -+ P is surjective and p : P + M is a right inverse for (Y, then p = p o (Y is an idempotent endomorphism of M, since (P 0 42 = (P O a) O (P O a) =po(cyop)ocr =/30idpocx=Pocx, and then 2 H ((Y(Z), (l-p)(z)) g ives an isomorphism M E P@(l-p)(M). Using this observation, we can now prove the fundamental characterization of projective modules. 1.1.2. Theorem. Let R be a ring. An R-module is projective if and only if it is isomorphic to a direct summand in a free R-module. It is finitely generated and projective if and only if it is isomorphic to a direct summand in R” for some n. Proof. If P is projective, choose a free module F and a surjective R-module homomorphism (Y : F -+ P by taking F to be the free module on some

1. Defining Ko

3

generating set for P, and (Y to be the obvious map sending a generator of F to the corresponding generator of P. We are using the universal property of a free module: To define an R-module homomorphism out of a free module, it is necessary and sufficient to specify where the generators should go. If P is finitely generated, then F will be isomorphic to R” for some n. The observation above then shows P is isomorphic to a direct summand in a free R-module, which we can take to be R” for some n if P is finitely generated. For the converse, observe first that free modules F are projective, since given a surjective R-module homomorphism CY : M + F with F free, one can for each generator pi of F choose some yi E M with a(yi) = zi, and then one can define a right inverse to (Y by using the universal property of a free module to define an R-module homomorphism p : F + M with ~(zc,) = yi. Next, suppose F = P @ Q and F is a free module. Given a surjective R-module homomorphism a : M -+ P, Q! @ idQ is a surjective R-module homomorphism (M @ Q) + (P @ Q) = F, so it has a right inverse. Now restrict this right inverse to P and project into M to get a right inverse for CL Finally, if F = R” with standard generators ~1, . . . , x,, then P is generated by p(zi), where p is the identity on P and 0 on Q. Thus a direct summand in R” is finitely generated and projective. 0 We’re now almost ready to define KO of a ring R. First of all, note that the isomorphism classes of finitely generated projective modules over R form an abelian semigroup Proj R, in fact a monoid, with @ as the addition operation and with the O-module as the identity element. To see that this makes sense, there are a few easy things to check. First of all, Proj R is a set! (This wouldn’t be true if we didn’t take isomorphism classes, but in fact we have a very concrete model for Proj R as the set of split submodules of the R”, n E N, divided out by the equivalence relation of isomorphism.) Secondly, direct sum is well defined on isomorphism classes, i.e., if P E P’ and Q E Q’, then P@Q 2 P’@Q’. And thirdly, direct sum is commutative (P@QgQ@P) and associative ((P CD Q) @ V Z P @ (Q CB V)) once we pass to isomorphism classes. In general, though, Proj R is not a group, and may not even have the cancellation property a+b=c+b*a=c. It’s therefore convenient to force it into being a group, even though this may result in the loss of some information. The idea of how to do this is very simple and depends on the following, which is just a generalization of the way Z is constructed from the additive semigroup of positive integers, or Q” is constructed from the multiplicative semigroup of non-zero integers, or a ring is “localized” by the introduction of formal inverses for certain elements. 1.1.3. Theorem. Let S be a commutative semigroup (not necessarily having a unit). There is an abelian group G (called the Grothendieck group or group completion of S), together with a semigroup homo-

4

1. KIJ of Rings

morphism ‘p : S + G, such that for any group H and homomorphism $I : S -+ H, there is a unique homomorphism 6 : G + H with 11, = 8 o cp. Uniqueness holds in the following strong sense: if cp’ : S --t G’ is any other pair with the same property, then there is an isomorphism (I! : G + G’ with ‘p’ = a: o ‘p. Proof. We will outline two constructions. The simplest is to define G to be the set of equivalence classes of pairs (2, y) with 5, y E S, where (IC, y) N (u, U) if and only if there is some t E S such that (1.1.4)

z+v+t=u+y+t

ins.

Denote by [(CC, y)] the equivalence class of (2, y). Then addition is defined by the rule [(xc, Y)l + w, Y’)l = [(x + 2’7 Y + Y’N *

(It is easy to see that this is consistent with the equivalence relation, and that the associative rule holds.) Note that for any x and y in S,

[(XT x)1

= KY, Y)l

since x + y = y + 2. Let 0 be this distinguished element [(z, x)]. This is an identity element for G, i.e., G is a monoid, since for any x, y, and t in S,

(x + 6 Y + 4 - (xc, Y>. Also, G is a group since

[(XT Y)] + [(Y, x>l = Kx + Y1 2 + Y)l = 0. We define cp : S --+ G by

and it is easy to see that this is a homomorphism. Note that the image of ‘p generates G as a group, since

1(x, Y)l = (P(x) - cp(Y> in G. Given a group H and homomorphism $ : S + H, the homomorphism 0 : G --+ H with $J = 8 o cp is defined by

0 ([(x7 Y)l) = +,(x:) - q(Y). Alternatively, one may define G to be the free abelian group on generators [xl, x E S, divided out by the relations that if x + y = z in S, then the elements [x] + [y] = [z] in G. Note that [(x, y)] in the previous construction corresponds to [x] - [y] in this second construction. The map

1. Defining Ko

5

is 2 H [z], and of course any homomorphism from S into a group H must factor through G by construction. To prove the uniqueness, suppose ‘p’ : S + G’ has the same universal property. First of all, v’(S) must generate G’, since otherwise, if G” is the subgroup generated by the image of cp’, then there are two homomorphisms 0 : G’ -+ G’ @ G’/G” with cp

(9’7 0) = 0 0 4, namely, 13 = (id, 0) and 0 = (id, Q), Q the quotient map. By the universal properties for G and G’, there must be maps a : G + G’ with ‘p’ = (Y o cp and ,S : G’ + G with ‘p = /3 o cp’. But then Q: o p = id on the image of cp’, hence on all of G’, so (Y is a left inverse to p. Similarly p o (Y = id on the image of cp; hence o is also a right inverse to S, as required. Cl Remarks. The assignment S -+ G = G(S) is in fact a functor from the category of abelian semigroups to the category of abelian groups, since if y : S + S’ is a homomorphism of semigroups, it induces a commutative diagram S’ S A ‘p

1

9’

1

G(S) - G(W, where the arrow at the bottom is uniquely determined by the universal property of G(S). In fancier language, Theorem 1.1.3 just asserts that the forgetful functor F from the category of abelian groups to the category of abelian semigroups has a left adjoint, since Homsemisroups (S, FH) g HomGroups(G, H). This could also have been deduced from the adjoint functor theorem (see [Freyd] or [Mac Lane]). It is convenient that we do not have to assume that cancellation (z+z = y + z =+- x = y) holds in S. Indeed, the map ‘p : S -+ G is injective if and only if cancellation holds in S. One of the reasons for introducing Grothendieck groups is that semigroups without cancellation are usually very hard to handle; yet in many cases their Grothendieck groups are fairly tractable. 1.1.5. Definition. Let R be a ring (with unit). Then Ko(R) is the Grothendieck group (in the sense of Theorem 1.1.3) of the semigroup Proj R of isomorphism classes of finitely generated projective modules over R. Note that Ko is a functor; in other words, if ‘p : R + R’ is an Rmodule homomorphism, there is an induced homomorphism Ko(cp) = cp* : Ko(R) + Ko(R’) satisfying the usual conditions id, = id, ((POT/J), = (P+o$J,. To see this, observe first that cp induces a homomorphism Proj R 4 Proj R’

6

1. KO of Rings

via [P] H [R’ @+, P], for P a finitely generated projective module over R. As required, R’ @p P is finitely generated and projective over R’, since if P@Q 2 R”, then (R’ cg~~ P) $ (R' c&, Q) 2 R' EQ, (P @Q)

z (R’

&, Rn) =

R’“.

And of course, the tensor product commutes with direct sums so we get a homomorphism. F’unctoriality of Ke now follows from functoriality of the Grothendieck group construction. 1.1.6. Example. If R is a field, or more generally a division ring (i.e., a skew-field), then any finitely generated R-module is a finitely generated R-vector space and so has a basis and a well-defined dimension. This dimension is the only isomorphism invariant of the module, so we see that Proj R E N, the additive monoid of natural numbers. Since the group completion of N is Z, &(R) Z Z, with the isomorphism induced by the dimension isomorphism Proj R -t N. The inclusion of a field F into an extension field F’ induces the identity map from Z to itself, since dimF,(F’ @F P) = dimF P for any F-vector space P. This same example also shows why we only use finitely generated projective modules in defining Ke. If R is a field, the same arguments show that the monoid of isomorphism classes of countably generated modules is isomorphic to the extended natural numbers N U {oo), with the usual rule of transfinite arithmetic, n + 00 = 00 for any n. This is no longer a monoid with cancellation; in fact, any two elements become isomorphic after adding 00 to each one. Thus the Grothendieck group of this monoid is trivial. A similar phenomenon happens with infinitely generated modules over an arbitrary ring; see Exercise 1.1.8. 1.1.7. Exercise. Let S be the abelian monoid with elements a,, m, where nfN,and m=Oifn=Oor 1, mEZifn=2, ( m E Z/2 if n 2 3. The semigroup operation is given by the formula

a,, m +

ad, d

=

h+d,

m+m’,

where m + m’ is to be computed in Z if n + n’ 5 2 and in Z/2 if n + n’ _> 3. (If for instance n = 2 and n’ > 1, then m is to be interpreted mod 2.) We shall see in $1.6 that S is isomorphic to Proj R with R = CR(S2), the continuous real-valued functions on the 2-sphere. Compute G(S) and the map cp : S --) G(S). Determine the image of S in G, and show that while cp-‘(0) = 0, ‘p is not injective.

2. KO from idempotents

7

1.1.8. Exercise (the “E ilenberg swindle”) . Show that for any ring R, the Grothendieck group of the semigroup of isomorphism classes of countably generated projective R-modules vanishes. 1.1.9. Exercise. Recall that if a ring R is commutative, then every left R-module is automatically a right R-module as well, so that the tensor product of two left R-modules makes sense. (1) Show that the tensor product of two finitely generated projective modules is again finitely generated and projective. (2) Show that the tensor product makes KC,(R) into a commutative ring with unit. (The c 1ass of the free R-module R is the unit element .)

2. KO from idempotents There is another approach to Ke which is a little more concrete and therefore often convenient. If P is a finitely generated projective R-module, we may assume (replacing P by an isomorphic module) that P @ Q = R” for some n, and we can consider the R-module homomorphism p from R” to itself which is the identity on P and 0 on Q. Clearly p is idempotent, i.e., p2 = p. Since any R-module homomorphism R” -+ R” is determined by the n coordinates of the images of each of the standard basis vectors, it corresponds to multiplication on the right (since R is acting on the left) by an n x n matrix. In other words, P is given by an idempotent n x n matrix p which determines P up to isomorphism. On the other hand, different idempotent matrices can give rise to the same isomorphism class of projective modules. (When R is a field, the only invariant of a projective module P is its dimension, which corresponds to the rank of the matrix p.., When the characteristic of the field is zero, the rank of an idempotent matrix is just its trace.) So to compute Ko(R) from idempotent matrices, we need to describe the equivalence relation on the idempotent matrices that corresponds to isomorphism of the corresponding modules. 1.2.1. Lemma. If p and q are idempotent matrices over a ring R (of possibly different sizes), the corresponding finitely generated projective Rmodules are isomorphic if and only if it is possible to enlarge the sizes of p and q (by adding zeroes in the lower right-hand corner) so that they have the same size N x N and are conjugate under the group of invertible N x N matrices over R, GL(N, R). Proof. The condition is sufficient since if u E GL(N, R) and upu-i = q, then right multiplication by u induces an isomorphism from RNq to RNp. So the problem is to prove necessity of the condition. Suppose p is n x n and q is m x m, and R”p % Rmq. We can extend an isomorphism CY : Rnp + Rmq to an R-module homomorphism R” + R” by taking (Y = 0 on the complementary module R”( 1 - p), and by viewing the image Rmq

8

1. KO of R i n g s

as embedded in R”. Similarly extend oy-’ to an R-module homomorphism ,0 : R” + R” which is 0 on R”( 1 - q). Once we’ve done this, cr is given by right multiplication by an n x m matrix a, and p is given by right multiplication by an m x n matrix b. We also have the relations ab = p, ba=q,a=pa=aq,b=qb=bp. ThetrickisnowtotakeN=n+mand to observe that lip l:q)2=(; ( (with usual block matrix notation) and that ( lip

laq)(l;

:)(lbp =

;)

1:q) ( lip

I:,)(:

;)=(:

:).

T h u s ‘-’ a is invertible and conjugates p@O to O@q. The latter l-q > ( b is of course conjugate to q @ 0 by a permutation matrix. 0 Now we can give a simple description of Proj R. 1.2.2. Definition. Let R be a ring. Denote by M(n, R) the collection of n x n matrices over R and by GL(n, R) the group of n x n matrices over z i (this is a non( ) unital ring homomorphism) and GL(n, R) in GL(n + 1, R) by the group a 0 h o m o m o r p h i s m a I+ o 1 . Denote by M(R) and GL(R) the infinite ( ) unions of the the M(n, R), resp. GL(n, R). Note that M(R) is a ring without unit and GL(R) is a group. It is important to remember that each matrix in M(R) has finite size. Let Idem be the set of idempotent matrices in M(R), and note that GL(R) acts on Idem by conjugation. R. We embed M(n, R) in M(n + 1, R) by a I+

Now we can restate Lemma 1.2.1. 1.2.3. Theorem. For any ring R, Proj R may be identified with the set of conjugation orbits of GL(R) on Idem( The semigroup operation is : i . (Oneonlyh as commutativity and associa( > tivity after passage to conjugacy classes.) Ko (R) is the Grothendieck group of this semigroup.

induced by (p, q) I-+

Using this fact we can now show that Ke is invariant under passage from R to M,(R) and commutes with direct limits. We will also construct an example of a ring for which Ko vanishes. 1.2.4. Theorem (“Morita invariance”). For any ring R and any positive integer n, there is a natural isomorphism Ko(R) 2 Ko(M,(R)). Proof. Via the usual identification of Mk(Mn(R)) with Mkn(R), Idem(M,(R)) = Idem

and GL(M,(R)) = GL(R).

2. KO from idempotents

The result therefore follows immediately from Theorem 1.2.3.

9

0

Next we show that Ko is a continuous functor, i.e., that it commutes with (direct) limits. A direct system or directed system in a category is a collection (Aa)aE~ of objects, indexed by a partially ordered set I with the property that if (Y, 0 E I, there is some y E I with y 2 CX, y > p. In addition, one supposes there are morphisms (~~0 : A, + Ag defined whenever a 5 /3, with the compatibility condition ‘PBr o ‘PM = (Pa-i, o = Ko(R)/L,(Z). Note that we have seen that &(R) vanishes if R is a division ring or a PID. In general, l&,(R) measures the non-obvious part of Ko(R). We will see in the next section that it recaptures a famous classical invariant of Dedekind rings. Next we compute Ke for local rings (which are not necessarily commutative). We begin with a review of some useful general ring theory. 1.3.3. Definition. A ring R (not necessarily commutative) is local if the non-invertible elements of R constitute a proper 2-sided ideal M of R. Examples of commutative local rings include k[[t]], the ring of formal power series over a field k, and Zc,), the ring of rational numbers of the form %, where p is a prime, b # 0, and p { b. For an example of a non-commutative local ring, let S be any non-commutative unital k-algebra, where k is a field, and let R = {a, + alt + a2t2 + . . . E S[[t]] : ao E k} . Since any power series in R with a0 # 0 is invertible (by the usual algorithm for inverting power series), and since the elements in R with ae = 0 constitute an ideal, R is a local ring. 1.3.4. Proposition. For a ring R (not necessarily commutative), the following are equivalent: (a) R has a unique maximal left ideal, and a unique maximal right ideal, and these coincide. (b) R is local. Proof (b) + (a). If R is local with ideal M of non-invertible elements, no element of R \ M can lie in a proper left ideal or proper right ideal, hence M is both the unique maximal left ideal and the unique maximal right ideal. Now let’s show (a) + (b). Assume (a) and let z E R. If z does not have a left inverse, then Rx is a proper left ideal, which by Zorn’s Lemma lies in a maximal left ideal, which by (a) is unique. Similarly, if x does not have a right inverse, then x lies in the unique maximal right ideal. Thus all non-invertible elements lie a proper 2-sided ideal M. 0

3. Ko of PIDs and local rings

13

1.3.5. Corollary. In a local ring, an element with a one-sided inverse is invertible. Remark. Note that replacing (a) by the condition that R has a unique maximal 2-sided ideal gives a very different class of rings in the noncommutative case. A simple ring R (one with no 2-sided ideals other than 0 and R) need not be local; a matrix ring over a field is a counterexample, since a sum of singular matrices need not be singular. 1.3.6. Definition. If R is any ring, the radical (or Jacobson radical) of R is the intersection of the maximal left ideals. By Proposition 1.3.4, in a local ring, the radical coincides with the maximal ideal. 1.3.7. Proposition. For any ring R, the radical of R is a 2-sided ideal. Proof. If I is a maximal left ideal, the annihilator of R/I in R certainly is contained in I. Hence

n

Anna(R/I) 2 (7 I = rad R.

I a max. left ideal

I

On the other hand, Anna(R/I)

=

n

Anna(k),

&R/I,i#O an intersection of maximal left ideals. So rad R is exactly the intersection of the AnnR(R/I), and so is 2-sided. 0

Remark. The proof showed that the radical of R is the set of elements that annihilate all simple left R-modules. One observation we will need later is that since every simple module for M,(R) is isomorphic to one of the form R” @R M with M a simple R-module, any matrix all of whose entries lie in rad R must annihilate all such modules, hence must be in the radical of M,(R). 1.3.8. Proposition. For any ring R, the radical coincides with {x E R : Va E R, 1 - az has a left inverse} and with the intersection of the maximal right ideals. Proof. First we show that rad R is contained in the indicated set. If x lies in every maximal left ideal, then Rz lies in every maximal left ideal. Suppose a E R and 1 - az does not have a left inverse. Then 1 - ax lies in a proper left ideal, hence in a maximal left ideal M. Since ax E M, we have 1 E M, a contradiction. Conversely, suppose that for all a E R, 1 - aa: has a left inverse. Let M be a maximal left ideal. If x 4 M, then Rx + M = R. Thus for some a E R, 1 - ax E M, a contradiction. So rad R coincides with {x E R : Va E R, 1 - ax has a left inverse}.

14

1. KO of Rings

Similarly, we can define the right radical r-4 R = n max. right ideals = {z E R : Va E R, 1 - xa has a right inverse}. Since rad R is a right ideal by 1.3.7, if x E rad R and a E R, there is a c E R with (1 - c)(l - xa) = 1. This gives (1 - xa)(l - c) = 1 + zac - cxa, and since x E rad R, xac - cxa E rad R. Thus 1 + xac - cxa has a left inverse, which shows 1 -c has a left inverse. Since it also has a right inverse, namely 1 - xa, they coincide, and 1 - za is invertible with inverse 1 - c. Hence rad R s r-rad R. By symmetry, r-rad R E rad R and the two coincide. El 1.3.9. Theorem (Nakayama’s Lemma). Suppose R is a ring and M is a finitely generated R-module such that (rad R)M = M. Then M = 0. Proof. Suppose M # 0. Pick a set of generators x1, . . . , x, for M with m as small as possible. (This implies in particular that each xj # 0.) By the assumption that (rad R)M = M, there are elements ~1, . . . , r, in rad R such that Hence (1 - T,)X, = ?-1x1 + -. . + Tm-_1Xm__1. By Proposition 1.3.8, 1 - T, is invertible; hence x,,, can be expressed as a linear combination of x1, . . . , x,-l. This contradicts the assumption that m was as small as possible. 0 1.3.10. Corollary. If R is a ring, M is a finitely generated R-module, andxl,..., xm E M, thenxl,... , x, generate M if and only if their images &, . . . , &,, generate M/(rad R) M as an R/ rad R-module. Proof. The “only if” statement is trivial. Suppose 21, . . . , km generate M/(rad R)M. Let iV = &l + . . . + Rx, C M and consider M/N. This satisfies the hypotheses of Nakayama’s Lemma, so M/N = 0 and M = N. 0 1.3.11. Theorem. If R is a local ring, not necessarily commutative, then every projective finitely generated R-module is free with a uniquely defined rank. In particular, Ko(R) g Z with generator the isomorphism class of a free module of rank 1. Proof. Note R/ rad R is a division ring D. If M is a finitely generated projective R-module, we may assume M @ N = R” for some k. Then M/(rad R)M and N/(rad R)N are D-modules, hence are free, say of ranks m and n, respectively, with m+n = k. Choose basis elements and pull them back to elements xl,. . . ,x, E M, x,+1,. . . , xk E N. By Corollary 1.3.10, these generate R”. We want to show that xl,. . . , xk are a free basis for Rk . This will show in particular that x1, . . . , x, are a linearly independent generating set for M, so that M is free with the uniquely determined rank rank M = dimD M/(rad R)M.

3. Ko of PIDs and local rings

15

Let el,. . . , ek be the standard free basis for R”. Since we now have two generating sets for Rk, each can be expressed in terms of the other, and there are elements aij, bij E R with k

ei =

c

k

aijxj,

Xi =

j=l

Thus we get

so

c

bij ej .

j=l

k

k

j=l

kl

k

k(aijbjl - &)el = 0, j=l 1~1 and if A = (aij), B = (bij), this means (since the el are linearly independent) that AB = I. Substituting the other way, we get k

k

and since the ~1 are linearly independent modulo the radical of R, this shows BA - I E M,(rad R) E rad M,(R) (using the remark following 1.3.7). By Proposition 1.3.8, BA is invertible, hence B is invertible. Since A was a left inverse for B, this shows it is also a right inverse, i.e., BA = I. This proves the x1, . . . , x, are a free basis for R”. Cl Part of the interest in local rings stems from the importance of localization as a technique for studying more general commutative rings. Recall that if R is a commutative ring, the set Spec R of prime ideals in R becomes a topological space, called the spectrum of R, when equipped with the s+called Zariski topology. The closed sets EI in this topology are parameterized by the ideals I of R, where for I 9 R,

1.3.12. Proposition. Let R be a commutative ring and let Spec R be its prime ideal spectrum. If P is a finitely generated projective R-module, then P has a well-defined rank function rank P : Spec R -+ N, and this function is continuous. In particular, if R is an integral domain, it is constant. Furt_hermore, for any commutative ring R, there is a splitting &(R) = Z @ K,,(R). Proof. Given p E Spec R, Pp 2 R, @R P is a finitely generated projective module over R, , which is a local ring. So by Proposition 1.3.11, it is free with a well-defined rank, which is the dimension of the associated module

16

1. Ko of Rings

over the field RP/mPl where mp is the unique maximal ideal of R,. Since mp = R, 8.R p, the rank at p may also be computed by first taking P/pP, which is a module over the integral domain R/p, then taking the dimension of the associated vector space over the field of fractions of R/p. Next we prove continuity of the rank function. One way of seeing this is via the idempotent picture. Suppose P is defined by an idempotent matrix p E M,(R). Then rankp P = k if and only if the image of p in M,(R/p) has rank k. Thus rankp P < k if and only if every (k + 1) x (k + 1) submatrix of p has a determinant in p. This is clearly a closed condition, since it’s equivalent to saying p contains certain specific elements of R, and the most general closed set in Spec R is of the form {p : p > I} for some ideal I. But it’s also an open condition since rankpskwrank(l-p)>n-k. To prove the final remarks, note that if R is an integral domain, then (0) is an open point in Spec R, hence Spec R is connected and rank P must be constant. The splitting map &Lo(R) -t Z for a general commutative ring is obtained simply by fixing a point p E Spec R and computing the rank there. Cl 1.3.13. Exercise (The finite generation hypothesis in Nakayama’s Lemma is necessary). Show from Nakayama’s Lemma that if R is a left Noetherian ring and (rad R)2 = rad R, then rad R = 0. Let R be the ring of germs at 0 of continuous functions I[$ + Iw. Show that R is a local ring, with radical the germs of functions f with f(0) = 0, and that (rad R)2 = rad R. (R is not Noetherian, which is why this is possible.) 1.3.14. Exercise. Compute Ke(Z/(m)) in terms of m, for any integer m > 0. Hint: write m aa a product of prime powers and use the Chinese Remainder Theorem to get a corresponding splitting of Z/(m) as a product of local rings. Then use Theorem 1.3.11 and Exercise 1.2.8.

4. KO of Dedekind domains A particularly rich family of rings for which Ke is interesting are the Dedekind domains. We begin with the definition and basic properties of these domains, and then proceed to the most important examples, namely, the rings of integers in number fields. In this section R will always denote a commutative integral domain embedded in its field of fractions F. 1.4.1. Dekinition. A non-zero R-submodule I of F is called a fractional ideal of R if there exists some a E R with aI C R. Clearly a non-zero ideal of R may be viewed as a fractional ideal; for emphasis, such an ideal is called an integral ideal. Also, if 5 E F (a, b E R; a, b # 0), then R( f ) is a fractional ideal since bR( %) C R. Such a fractional ideal is called

4. Ko of Dedekind domains

17

principal. One can multiply fractional ideals, and under multiplication they form an abelian monoid with identity element R. 1.4.2. Definition. R is called a Dedekind domain or Dedekind ring if the fractional ideals under multiplication are a group, i.e., if given a fractional ideal I, there is a fractional ideal 1-l with I-l1 = R. Observe that necessarily 1-l = {u E F : a1 C R}. For if J = {a E F : al z R}, then 1111 c R so II1 & J, but then R = II-’ C IJ C R, so II-l = IJ and II1 = I-‘I J = J. 1.4.3. Definition. Note that the principal fractional ideals are a subgroup of the fractional ideals isomorphic to FX/RX. The class group of the Dedekind domain R is defined to be C(R) =

{group of fractional ideals}/{group of principal fractional ideals}. 1.4.4. Proposition. The class group of a Dedekind domain may also be identified with the set of R-module isomorphism classes of integral fractional ideals. Proof. Clearly any fractional ideal is isomorphic to an integral one I (via multiplication by some element of R \ (0)). And if I = (J)(Ra), then multiplication by a implements an R-module isomorphism J -+ I. Conversely, if cp : I -+ J is an R-module isomorphism and ue E I 1 {0}, then for any a E I,

so cp(ue)l = ueJ and [I] = [J] in C(R).

Cl

1.4.5. Theorem. If R is Dedekind, then every fractional ideal is finitely generated and projective. In particular, R is Noetherian. Proof. Let I be a fractional ideal. Since II11 = R, there are elements E 1-l and ~1,. . . , yn E I such that x:=1 ziyi = 1. If b E I, then b = C(bxi)yi with bxi E I-‘I = R, so ~1,. . . , yn generate I. Thus I is finitely generated. Since every ideal of R is finitely generated, R is Noetherian. But in addition, the homomorphism R” --f I defined by (ai, . . . , a,) I+ C aiyi splits, with right inverse b H (bxl, . . . , bxn), by the same calculation. So I is isomorphic to a direct summand in R” and so is projective by Theorem 1.1.2. 0

n,...,Gx

1.4.6. Corollary. If R is Dedekind, then every finitely generated projective R-module is isomorphic to a direct sum of ideals. In particular, the isomorphism classes of the ideals generate Ke (R) . Proof. We use the same argument as in the proof of Theorem 1.3.1. Let A4 be a finitely generated projective module over R. We may assume

18

1. Ko of R i n g s

that M is embedded in some Rn. We argue by induction on n that M is isomorphic to a direct sum of k ideals for some k 5 n. If n = 0, there is nothing to prove. So assume the result for smaller values of n and let r : R” ---) R be projection on the last coordinate. Note that n maps M onto an R-submodule of R, i.e., an ideal. If r(M) = 0, then we may view M as embedded in kerr z R”-’ and use the inductive hypothesis. Otherwise, n(M) is a non-zero ideal I and so is projective by Theorem 1.4.5. Hence M splits as kern\M @ I (recall the remarks in 1.1.1). Since we may view kern)M as embedded in Rnpl, we may apply the inductive hypothesis to conclude that it’s isomorphic to a direct sum of k’ ideals, k’ < n - 1. So M is a direct sum of k ideals with k = k’ + 1 2 (n - 1) + 1 = n. Cl Our next goal is to relate Ko(R) to C(R), but first we need to develop more of the theory of Dedekind domains. This will also enable us to prove a useful characterization of Dedekind domains that will show that the ring of algebraic integers in a number field is a Dedekind domain. The next theorem generalizes the “fundamental theorem of arithmetic” (unique factorization of an integer into primes). 1.4.7. Theorem. In a Dedekind domain R, every prime integral ideal is maximal. And every proper integral ideal can be factored uniquely (up to renumbering of the factors) into prime (or maximal) ideals. The group of fractional ideals is the free (multiplicative) abelian group on the (non-zero) prime ideals. Proof. (a) Suppose 0 5 I 5 R and I is prime but not maximal. Then there exists an integral ideal J with I 5 J 5 R. Let K = J-lI; since I 5 J, K 5 J-l J = R. Since JK = I and I is prime but J g I, K C I. But then I = JK C JI 5 RI = I, a contradiction. So I is maximal. (b) Existence of factorizations. Let C = {proper integral ideals that are not products of prime ideals}. If this is empty, we’re done. Otherwise, since every ascending chain of ideals in R has a maximal element (R is Noetherian by Theorem 1.4.5), C has a maximal element I by Zorn’s Lemma. I can’t be a maximal ideal (otherwise it would be prime itself and would have a trivial factorization I = I) so I 5 Ii 5 R for some ideal 11. Let 1, = lcll. This is also an ideal in R since I C II, and since I 5 II, it is a proper ideal containing I properly. Since 11 and 1s are both strictly bigger than I and I was maximal in C, both have factorizations into primes. But since I = 1112, multiplying gives a factorization of I, a contradiction. (c) Uniqueness of factorizations. Suppose Pi .. . P, = &I . . . Qn with Pi, Qj prime and m < n. Then PI > PI. .. P, = Q1 . . .Qn so some Qj lies in PI. After renumbering if necessary, we may assume Qi C PI. Write Qi = SlPl by the Dedekind property (where 5’1 = Pc’Q1). Multiplying through by PF1 gives Ps . . . P, = SlQ2 . . . Qn. Continuing by induction, we get down to the case where m = 1, in which case it is clear

4. Ko of Dedekind domains

19

that we must have n = 1 and Qr = PI. So factorizations into primes are unique. (d) Clearly there’s a map from the free abelian group on the prime ideals into the multiplicative group of the fractional ideals. By (b) above, it’s surjective. If there is something non-trivial in the kernel, we have P”‘.. . P,“? = R for some distinct prime ideals Pj and some nj E Z. If for szme j, nj < 0, multiply through by P!““ . Then we end up with some ideal in R having two distinct factorizat:ons, contradicting (c). Cl 1.4.8. Lemma. Let R be any commutative ring and let Ii, 1s be ideals inR. IfIl+Is=R, thenIiIs=IInIs. Proof. Clearly 111s C Ii n1s. On the other hand, if al E 11, a2 E 12, and al + a2 = 1, then for x E Ii n 4, x = alx + asx E 1112 + 1211 = Ii&. 0 1.4.9. Lemma. Let R be a Dedekind domain and let I be a fractional ideal, J an integral ideal. There exists a E I such that I-la + J = R. Proof. Let PI,. . . , P, be the distinct prime ideals that occur in the factorization of J given by Theorem 1.4.7. Choose ai E IPi . + . i)i.. . P,. with ai 4 IPi e.1 P,.. Let a = cai. Note aiI_’ C Pj if j # i, but aiI_’ g Pi, site otherwise we’d have ail-l cnPj=Pl” . P,

by iterated use of (1.4.8),

hence ai E IPl*..Pr, a contradiction. Now note that I-la p Pj for any j. It’s an integral ideal and this says I-la+ J can’t be divisible by any Pj. But it can’t be divisible by any other prime ideal, either, by the choice of a, so it can’t be a proper ideal and must be all of R. Cl This implies that a Dedekind domain doesn’t miss being a PID by very much. If R is a PID, any fractional ideal is singly generated. In a Dedekind domain, the best one can say along these lines is the following. 1.4.10. Corollary. If R is a Dedekind domain, any fractional ideal of R can be generated by at most two elements. Proof. Let I be a fractional ideal, 0 # b E I. Let J = bI_‘, which is an integral ideal. By Lemma 1.4.9, there is some a E I with al-’ + bI_l = R. ThenI=Ra+Rb. 0 1.4.11. Lemma. Suppose R is a Dedekind domain and II, I2 are fractional ideals for R. Then Ii 03 I2 2 R 03 IiI2 as R-modules. Proof Choose al # 0 in 11 and let J = alI;‘, which is an integral ideal. Apply Lemma 1.4.9 with I = Is. We get a2 E Is such that I;‘as+alI;’ = R. Choose bl E IL1, b2 E Ipl with albl + a2b2 = 1. Then

20

1. KIJ of Rings

showing that (ii Ly) is invertible with inverse ( :i,

I:), and

(21, $2) H (21, x2)

gives the desired isomorphism (with inverse given by multiplication by the inverse matrix). Cl 1.4.12. Theorem. Let R be a Dedekind domain. Then any projective R-module of rank k can be written as Rkml ~3 I, with I an ideal, and the isomorphism class of I is uniquely determined. If P and Q are finitely generated projective modules of the same rank k, say P Z R”- l $ I1 and Q Z R”- l@ 12 for ideals Ii and 12, the map [P] - [Q] I-+ IlIT1 sets up an isomorphism from go(R) to C(R). In fact,

[Rk-' @I] I-+

(k, [I])

sets up an isomorphism of abelian groups K,,(R) + Z @ C(R). As a commutative ring (see 1.1.9), Ko(R) = {(k, [I]) : k E z,

[I] E W-O),

with the operations (k, [I]) + (k’, [I’]) = (k + k’, W’]), (1.4.13)

(k, [I]) . (6 [1’1) = F’, [Il”‘[I’j”), rank : (k, [I]) ++ k E Z.

Proof. By Corollary 1.4.6, every finitely generated projective module P over R is isomorphic to a direct sum Ii @ . . . @ 4 of ideals, and by Proposition 1.3.12, P also has a well-defined rank. If I is an ideal, then rank I = dimF(F@R I) = dimF F = 1, so the rank of P is just the number k of ideals in a direct sum decomposition. Using Lemma 1.4.11 iteratively, we can rework the decomposition into the form Rkwl @I with a single ideal I. The only problem is to show that if (R”- ’ @ 11) 2 (R”- ’ @ 12), then Ir Z 12 as R-modules, or (equivalently, by Proposition 1.4.4) [II] = [I21 in C(R). 0 nce this is done, the formulae 1.4.13, and the identification of ko(R) with C(R), then follow upon taking the direct sum or tensor product of Rk @ I and of R”’ @ I’ and applying Lemma 1.4.11 iteratively.

4. KO of Dedekind domains

21

So suppose we have an isomorphism

with inverse p. Since any R-module map from one ideal to another is given by multiplication by an element of F (compare the proof of Proposition 1.4.4), a and p are induced by right multiplication by lc x k matrices A and B (with entries in F) which are inverses of each other. Now if X is the diagonal matrix with diagonal entries (1, 1, . . . ,1,x), where z E 11, then right multiplication by X maps R” into R”-’ @ II, hence right multiplication by XA maps Rk into Rkml @ 12. The rows of XA are the images of the standard basis vectors for Rk under this map, so they have their first k - 1 entries in R and last entry in 12. Thus expansion of the determinant along the last column shows that det(XA) E 1s. Since det X = 2, we obtain the condition z det A E I2 for all z E Ii. Similarly y det B = y(det A)-l E II for all y E 1s. So multiplication by det A implements an isomorphism from IitoIs. 0 We proceed now to the characterization of Dedekind domains. This will eventually make it possible to show that the rings of integers in number fields are Dedekind domains. Recall that a subring R of another ring S is called integrally closed in S if any element of S which is a root of a manic polynomial with coefficients in R actually lies in R. 1.4.14. Lemma. Let R be a Noetherian integral domain which is in& grally closed in its field of fractions F. Suppose I is a fractional ideal of R. Then {s E F : SI G I} = R. Proof. Since R is Noetherian, I is finitely generated. Let S = {s E F : SI s I}. Clearly R C S. But ifs E S, s is integral over R, by the following argument. Choose generators aj for I. Then there are elements bjk E R such that saj = c bjkak. Thus if B = (b.jk), s is an eigenvalue of B and so is a root of its characteristic polynomial, which is a manic polynomial with coefficients in R. Hence s E R since R is integrally closed. Thus SCR. • i 1.4.15. Lemma. Let R be a Noetherian commutative ring and let I be a non-zero proper ideal of R. Then I contains a product of non-zero prime ideals. Proof. Suppose the result is false, and let C be the family of non-zero proper ideals of R which do not contain a product of non-zero prime ideals. Since R is Noetherian, C must contain a maximal element (under inclusion), say I. Clearly I is not prime, so there must be a, b E R with ab E I, a, b $! I. We have I !j I + Ra, I $ I + Rb. If I + Ra = R, then (I+Ra)(I+Rb) = I+Rb 2 I, whileon theotherhand (I+Ra)(I+Rb) & I+Rab c I, a contradiction. So I 5 I+Ra 5 R. Similarly I !j I+Rb 5 R. Since 1 was maximal in C, I + Ra and I + Rb do not lie in C. Thus I+Ra> PI.. .Pr, Ii-Rb_>Ql.. . Qs, for some prime ideals Pj and Qk. Then I = (I + Ra)(I + Rb) 2 PI . . . P,Ql . . . Qs, a contradiction. 0

22

1. KO of Rings

1.4.16. Lemma. Let R be a Noetherian integral domain in which every prime ideal is maximal. Let I be a non-zero proper ideal of R. Then there exists c E F with c $ R such that cl C R. Proof. Let a # 0 in I. Then Ra contains by Lemma 1.4.15 a product of non-zero prime ideals, say PI . . . P,, and we may assume m is chosen to be minimal with this property. Let P be a maximal ideal containing I. Then PI . . . P, C Ra C I G P, so some Pj C P, say PI g P. Since all prime ideals are maximal, we have PI = P. If m = 1, then I = Ra = P is maximal and a-l $ R, a-‘I c R. If m 2 2, then by minimality of m, Ra 2 Pz . . . P,. Choose b E P2.. . P,,, withb$Ra,andletc=$. Thenc$!Rbut CI C_ cP1 = a-‘bPl c a-l PI . . . P, C_ a-l Ra = R.

Cl

1.4.17. Theorem. A commutative integral domain R is Dedekind if and only if it has the following three properties: (u) Every non-zero prime ideal is maximal. (b) R is integrally closed in its field of fractions F. (c) R is Noetherian. Proof. If R is a Dedekind domain, it satisfies (c) by Theorem 1.4.5 and (a) by Theorem 1.4.7. Suppose a E F, a # 0, and a is integral over R. Then a is a root of some manic polynomial ZP + an_~xn-’ +. . - + ao, where ao,*.., a,,_1 E R. Consider M = R + Ra + Ra2 + . +a + Ran-l. This is an R-submodule of F, and since an = -a,_la”-’ - ... - a~, it is stable under multiplication by a. If we write a = E, p, q E R and q # 0, then Q n-lM G R, so M is a fractional ideal. Multiplying aM E M by M-l gives aR s R, so a E R. This shows R is integrally closed. Now we show the conditions (a)-(c) imply R is Dedekind. Suppose R satisfies (a)-(c) and I is a fractional ideal. Let J = {a E F : aI E R}. W e want to show IJ = R, so that J is an inverse for I. Now IJ is an integral ideal. Let K = {a E F : aIJ E R}. By definition, K(IJ) = (KJ)I C R, so KJ c J. By Lemma 1.4.14, K 2 R. On the other hand, if IJ 5 R, then K 2 R by Lemma 1.4.16, a contradiction. So IJ = R and I is invertible. Cl 1.4.18. Theorem. Let F be a number field, i.e., a finite algebraic extension of Q, and let R be the ring of algebraic integers in F, that is, the integral closure of Z in F. Then R is a Dedekind domain. Proof. We need to check the conditions of Theorem 1.4.17. Condition (b) is the easiest. R C F, and if a E F is integral over R, then it is integral over Z by “transitivity of integrality,” hence already contained in R. So R is integrally closed. To check (a), let p be a non-zero prime ideal in R. Then p fl Z is a prime ideal in Z. We claim it can’t be zero. Indeed, if b # 0 is in p,

4. Ko of Dedekind domains

23

the product NQ(~J,Q(~) of the conjugates of b (in some Galois extension K 2 F) is f the constant term of the minimal polynomial of b, which by the assumption that b E R has coefficients in Z. Now this product of the conjugates of b is a product of b with a product c of other algebraic integers, and since bc E Z C F, c E F and is integral over Z. Hence c E R and 0 # bc E Rb n Z C p n Z. Thus p II Z is a non-zero prime ideal in Z, i.e., p fl Z = (p) for some prime number p. Since F is a finite algebraic extension of Q, R/p must be contained in a finite algebraic extension of V(n n z) = V(P), in other words in a finite field of characteristic p. Since a finite integral domain is a field, R/p is a field, i.e., p is a maximal ideal. It remains to check (c), i.e., that R is Noetherian. One way of seeing this is by using the trace. Recall that if z E F, T~F,Q(z) is the trace of the linear operator of multiplication by x on F, when we regard F as an n-dimensional vector space over Q, where n = [F : Q]. The trace pairing (2, Y) H %/Qh) is a non-degenerate symmetric Q-bilinear pairing on F (since for x # 0 in F, Tr ~/Q(xx-‘) = n # 0). Choose elements xi,. . . , A, E R which span F over Q. (One may obtain such elements by taking any basis elements for F over Q and then multiplying them by suitably large (ordinary) integers to kill off any denominators in the coefficients of their minimal polynomials.) Then x H (nF/Q(xh), . *. 7 %/Q&)) is an embedding of R into Zn. In particular, R is a finitely generated &module, so any ascending chain of ideals in R is an ascending chain of submodules in a finitely generated Z-module, and so terminates (since Z is Noetherian). Thus R is Noetherian. 0 Finally, we show that the Dedekind domains given by Theorem 1.4.18, which are the main subject of study in algebraic number theory, have finite class groups. The computation of these groups is not easy and is a problem of major interest. 1.4.19. Theorem. Let F be a number field, i.e., a finite algebraic extension of Q, and let R be the ring of algebraic integers in F, that is, the integral closure of Z in F. Then the class group I?-,(R) is finite.

Proof. The proof requires the notion of the norm of an ideal. If I is an integral ideal of R, with prime factorization PI”’ . . . Pm’, then by the Chinese Remainder Theorem,

Since R/Pj is a finite field for each j (by the proof of Theorem 1.4.18) and R/PJyi clearly has a composition series with nj composition factors, each isomorphic to R/Pj, R/PJJ’ is finite with IR/PjIn’ elements, and R/I is finite. Thus we can define

j=l

24

1. KO of Rings

It is clear that this norm is multiplicative: 11~1~211 =

11-7111 .

111211.

If I happens to be a principal ideal (a), note that since NF,q(a) is the determinant of the Z-linear operator of multiplication by a on R (which is isomorphic to Z” as a Z-module), Ra has index ( NF,Q(u)) in R and thus

II(~ = INF/&+ Recall from the proof of Theorem 1.4.18 that if P is a prime ideal with PI% = (p), then R/P is a finite extension of Z/(p) of degree 5 n = [F : Q], so that [[PII = pi for some j with 1 5 j 5 n. Thus for any C > 0, lIPI\ 5 C implies p 5 C for the corresponding p. On the other hand, for a fixed prime number p, there are only finitely many prime ideals P c R with PnZ = (p) (namely, those prime ideals occurring in the prime factorization of Rp). So putting all of this together, we see there are only finitely many ideals I satisfying 11111 < C. To prove the theorem, it therefore suffices to show that there is a constant C > 0 such that every element of C(R) has a representative I with 111/l 5 C. Choose a basis Xi,. . . , A, for R as a Zmodule. (That such a basis exists was shown in the proof of Theorem 1.4.18.) Let A be the maximal absolute value of a conjugate of one of the Xj in @ and let C = n”A”. Choose any element of C(R) and represent it by a fractional ideal of the form K = J-l, with J an integral ideal. We will show there is another representative I for the same ideal class with IlIll 5 C. Consider the set S = {oiXr + .. . + anAn : aj E 27, 0 5 aj 5 [II Jll’]*} (The square brackets denote the “greatest integer” function.) This set has ([llJll"l + 1)” > IIJII = IRIJI elements, so there must by the pigeonhole principle be two elements 17 and C of 5’ with the same image in R/J. Let c = q - C E J and let I = ( 1. Deduce that the class group kc(R) is generated by the classes of the ideals (Z - (Y, y-p), where (Y, p E lR ando2+p2=1. (3) Show that if pi and pz are prime ideals of the form (Z - oj, y - &), respectively, where oj, /?j E R and c$ +/3; = 1, j = 1, 2, then pip2 is a principal ideal, with generator a linear polynomial vanishing at both (~1, /?I) and ((~2, fi2), if these points are distinct, or else the linear polynomial oiz + /3iy - 1, if p2 = pi. Conclude that all non-principal prime ideals of R define the same element of the class group, and that this element is of order 2, hence that f?o(R) 2 Z/2. 1.4.24. Exercise (More on class groups of quadratic number fields). Let d be a square-free integer and let F = Q(a), which is the most general quadratic extension of Q. (1) Show that the ring R of algebraic integers in F is Z[Jd, provided

1

that d = 2 or 3 mod 4, and is Z @ if d E 1 mod 4. (This [ explains Exercises 1.4.20(l) and 1.4.2;(l).) (2) Let p E N be a (rational) prime. Show that R/(p) is a twodimensional algebra over the field F, of p elements, and that there are exactly three possibilities for R/(p): (4 R/(p) g bW(~“) contains a nilpotent element. In this case we say p is ramified. Show that this case happens exactly when p divides d or, if d G 2 or 3 mod 4, when p = 2. ( b ) R/(P) ” Fp 2 is a field, so the principal ideal (p) in R is maximal. In this case we say p is inert. (c) R/(P) = F, x F,. In this case we say the prime p splits in F. (Hint: suppose R = Z[c] with t2 = d, which is the case if d E 2 or 3 mod 4. Then R/(p) 2 IFP[z]/(z2 - d), so you just have to analyze whether the polynomial x2 -d has 0, 1, or 2 roots in Z/(p). The case d E 1 mod 4 is similar; it’s just that the polynomial is different.) 13) Show that in case (a), the ramified case, (p) = p2 for some prime ideal p of R, and that in case (c), the split case, (p) = plp2 for some distinct prime ideals of R. In either case, if R has no elements of norm p, then the prime ideals occurring cannot be principal and

5. Relative KO and excision

27

are thus non-trivial in C(R). Thus show that in the ramified case, one gets an element of C(R) of order 2. (4) Show how Exercise 1.4.20 fits into this general framework.

5. Relative KO and excision One of the things that makes K-theory so computable and useful is the fact that it behaves like a “homology theory” for rings. (The precise connection with a cohomology theory for topological spaces will be made in the next section.) In particular, when R is a ring containing a two-sided ideal I, there is an exact sequence relating Ke (R), Ko (R/I), and a certain “relative K-group.” This exact sequence looks something like the exact sequence in cohomology for a pair of topological spaces (X, A): Hj(X, A) --f Hj(X) -+ Hj(A). The first aim of this section is to define the relative group Ko(R, I) and the exact sequence relating it to K,-,(R) and Ko(R/I). Then we prove an algebraic analogue of the excision axiom for homology and develop some applications. 1.5.1. Definition. Let R be a ring and I C R an ideal (in this section, always two-sided). The double of R along I is the subring of the Cartesian product R x R given by D(R, I) = {(z, y) E R x R : 2 - y E I}. Note that if pi denotes projection onto the first coordinate, then there is a split exact sequence (1.5.2)

0 -+ I + D(R, I) = R + 0,

in the sense that pi is split surjective (with splitting map given by the diagonal embedding of R in D(R, I)) and that kerpr may be identified with I. 1.5.3. Definition. The relative Kc-group of a ring R and an ideal I is defined by K,,(R, I) = ker ((PI), : Ko(D(& 1)) -+ h(R)). Relative K-theory is closely linked to the phenomenon that while any matrix over R/I can be lifted to a matrix over R, an invertible matrix cannot always be lifted to an invertible matrix. The following lemma will also be used in the next chapter.

28

1. Ko of Rings

1.5.4. Lemma. Let R be a ring and I c R an ideal. Then if A E GL(n, R/I), the 2nx 2n matrix

(:

:I)

lifts to a matrix in GL(2n, R).

Proof. Note that (:

:I)=(:

:)(A-1

:)(:

:)(:

-b).

lifts “as is” to an invertible matrix over R. If B y _d ( > and C are any (not necessarily invertible) matrices in M,(R) lifting A and A-l, respectively, then The matrix

(:, :)

and

(:, :)

and

(-‘c ;)

are invertible and lift

y).

(-1-l

Now just multiply. Cl 1.5.5. Theorem. Let R be a ring and I c R an ideal. Then there is a natural short exact sequence

where q* is induced by the quotient map q : R + R/l and the m a p Ko(R, I) + I&(R) is induced byp2 : D(R, I) + R. Proof. For simplicity of notation in the proof, if A is an element of R or a matrix with entries in R, we will often denote q(A), the corresponding matrix over R/I, by A. First consider an element [e] - [f] E Ko(R, I), where e = (el, e2), f = (fi, f2) E Idem(D(R, I)). The image of [e] - [f]

koKa(R x R) g Ko(R) x Ko(R) (using (1.24) is ([ell -

[fl], [ez] - [fz]).

9* 0 (p2>*([el - [fl) = 9*([e21 - [f21> = L&21 - [S21, [el] - [fi] = 0 since by assumption [e] - [f] E ker(pl),.

whereas e, f E D(R, I), 21 = 62 and fi = f2. Thus

k21 - if21 = kil - [SI] = q*([el]

-

[fi]) =

But since

0.

Hence the image of the first map is contained in the kernel of the second. Now suppose e, f E Idem and q*([e] - [f]) = [k] - [f] = 0. Then 6 and f are stably equivalent, so for suitably large T, 1@ i, = q(e c+ ir) - j CB i, =

4(f a3 b)

5. Relative I(0 and excision

29

under GL(R/I). Replacing e by e $1, and f by f @ l,, we may assume i = &(@)-l for some matrix 4 E GL(R/I). In general, jl will not lift to a matrix in GL(R). However, jr@(i)-’ does conjugate &@b to f@& and lifts to a matrix h in GL(R) by Lemma 1.5.4. Thus we may replace f by f @ 0 and e by h(e CD O)h-l without changing [e] and [f, and reduce to the case where 1 = f. This means (e, f) E Idem(D(R, I)). Then [(e, e)] - [(e, f)] is a class in Ko (D(R, I)) which maps to 0 under (pi), and to [e] - [f under (~2)~. This completes the proof of exactness. The naturality of the sequence (under homomorphisms R + R’ sending I + I’) is obvious from the definition of the maps and from functoriality of Kc. Cl Remark. In general, the map &(R) + Ko(R/I) is not surjective, and the map &(R, I) --+ Ko(R) is not injective. The one exception will be the case where the ring homomorphism R -+ R/I splits. In this case it is obvious that the map Ko(R) --+ &(R/I) is split surjective, and it will also turn out (see 1.5.11 below) that Ko(R, I) is the kernel of this map. Next we want to prove the analogue of the excision theorem for top* logical homology. Recall that this says that under suitable hypotheses, the relative homology H.(X, A) is unchanged when a large subset U of A is removed from both A and X. Under optimal circumstances (for instance, for CW-pairs), H,(X, A) 2 &(X/A) and thus only depends on the “difference” between X and A. The analogous statement for Kc turns out to be true, and says that the relative group &(R, I) only depends on the “difference” between R and R/I, which is measured by I (with its structure as a ring without unit). In fact, it turns out that Kc makes sense and is functorial even for rings without unit and for non-unital ring homomorphisms. With this language, we show that &(R, I) g &(I). 1.5.6. Definition. Let I be a ring that doesn’t necessarily have a unit element. The ring obtained by adjoining a unit element to I, denoted I+, is as an abelian group just I @ Z, with multiplication defined by the rule (2, n) . (y, m) = (zy + ny + m2, mn), x, y E I; m, n E Z. It is an easy exercise to check that this is indeed a ring with unit, the unit element being (0, 1). The notation I+ is suggested by topology, where X+ is standard notation for a space X with a disjoint basepoint added. It is useful to note that if a! : I -+ I’ is a homomorphism in the category of rings without unit, it automatically extends uniquely to a unital homomorphism I+ 3 I:. Remark. The reader might wonder what happens if I already has a unit element, say e. In this case, there is a unital isomorphism o : I+ + I x Z (the Cartesian product of rings) defined by

a(~, n) = (x + ne, n),

30

1. Ko of Rings

since

a ((xc,

n>. (y, m)) = cr(sy + ny + mx, mn) =

(xy + ny + mx + mne, mn)

= (x + ne, n) + (y + me, m) = cx(2, n) . a(y, m).

1.5.7. Definition. Let I be a ring that doesn’t necessarily have a unit. Note that one has a split exact sequence (1.5.8)

0-+I+I+Gz+0.

Define &(I) = ker (p* : &(I+) + Ko(Z) E Z) . At first sight, there might appear to be some ambiguity here, since if I has a unit, we have given two different definitions of Ko(I). However, by the remark above, in this case I+ %! I x %, so &(I+) 2 I&(I) 6B Ko(Z), and ker p* just picks out the first summand. So the new definition agrees with the old one in this case. Also, this new definition makes & into a functor from the category of non-unital rings to abelian groups. This observation is occasionally useful even when one wants to deal only with rings with unit. For instance, if R is a ring with unit, there is a non-unital homomorphism R + M,(R) defined a 0 bya- o o . The reader can check that the homomorphism induced ( 1 by this non-unital homomorphism is the Morita invariance isomorphism of Theorem 1.2.4. 1.5.9. Theorem (Excision). If I is a two-sided ideal in a ring R, then Ko(R, I) G’ &(I) (and thus does not depend on R, only on the structure of I as a ring without unit). Proof. Define a unital homomorphism y : I+ + D(R, I) by (x,n)~(n~l,n~l+x),

xEI,

nEZ,

and note that the diagram

I+

r D(R,

I)

commutes. Hence y* : &(I+) -+ &(D(R, I)) sends kerp* to ker(p& i.e., maps &(I) to Ko(R, I).

5. Relative Ko and excision

31

Next we show that this map is surjective. Consider a class [e] - [f E Ko(& I), where e = (el, e2), f = (A, f2) E Idem(D(R, 1)) ad hl =

VII

in K,-,(R). After replacing e and f by eel, and f@l,. for a suitably large r, we may assume that ei and fi are conjugate under GL(R), say ei = gfig-’ for some invertible matrix g. Replacing (fi, f2) by (gfig-l, gfsg-‘), we may assume that in fact fi = ei. Next, if e is an s x s matrix, we may replace e and f by e @ (1, - ei, 1, - ei) and by f @ (1, - ei, 1, - ei). Note that there is an invertible 2s x 2s matrix h with entries in R conjugating el @(l, - el) to 1, @OS. Conjugating everything by h finally reduces us to the case where e = (ls @ O,, e2), f = (1, $ O,, f2). Since e and f are matrices over D(R, I), e2 - (ls @ 0,) and fs - (ls $0,) have entries in I. Now [e] - [f is clearly in the image of &(I). Finally, we have to show y* is injective on &(I). We may represent a general element of Ke (I) by [e] - [f , where e, f E Idem and rank p(e) = rank p(f). As above, if f is an r x T matrix, we may stabilize by taking direct sums with 1, - f and conjugating, and thus assume f = l,, rankp(e) = T. We may also assume gp(e)g-’ = 1, for some g E GL(Z). Viewing g as an element of GL(I+) via the split exact sequence 1.5.8, we may replace e by geg-’ and assume that p(e) = 1,. Now if +y,([e] - [l,.]) = 0, this means

[(L, e)l = [(b, L-11

in Ko(@R,

4).

We may stabilize if necessary by increasing r and assume that there is a matrix (gi, gs) E GL(D(R, I)) with giLg,l = lr,

g2eg;l = 1,.

Then (1, g;lgs) E GL(D(R, I)) and (g;1g2)e(g~1g2)-1

= gT1(g2eg,l)gl = gT1L-91 = 1,.

Since gL1g2 3 1 mod I, gy1g2 lies in GL(I+) and this says [e] - [lr] = 0 in Kc(l), proving that the kernel of y* is trivial. Cl 1.5.10. Examples. (a) Suppose R = Z and I = (m), where m > 0. Thus R/I = Z/(m). &(R/I) was computed in Exercise 1.3.14; the map Ko(R) --t &(R/I) is always injective but in general has a free abelian cokernel of rank k - 1, where k is the number of distinct prime factors of m. As a ring without unit, I is the free abelian group on a generator t satisfying t2 = mt. Hence I+ E iZ[t]/(t2 - mt), a fairly complicated ring. &(I) is not so easy to compute directly, though we will find a way to compute it in the next chapter. It turns out to be a finite abelian group. (b) For applications to topology (see Section 1.7 below), rings of the form R = ZG, the integral group ring of a group G, are of particular importance. It is a long-standing conjecture that when G is torsion-free, 20(R) = 0. This is known in some cases, for instance

32

1. Ko of Rings

when G is free abelian; this case will be treated in Chapter 3. For finite groups, Ke(ZG) is often non-trivial and contains interesting arithmetic information. Consider the simplest example, when G is cyclic of prime order p, say with generator t. Then R = ZG may be identified with Z[t]/(tp - 1). If E = e2rrilJ’, a primitive pth root of unity, and if S = Z[[], then S is the ring of integers in the cyclotomic field Q(c), hence is a Dedekind domain by Theorem 1.4.18. There is a surjective homomorphism R -+ S defined by sending t I-+ 5. Since the cyclotomic polynomial fp(t) = P-l + . . - + t + 1 is irreducible, any polynomial g(t) E Z[t] with g(E) = 0 must be divisible by f,. In particular, anything in the kernel I of the map R + S must be a multiple of fp. Note that as an element of R, fz = pfp. Thus I in this example is, as a ring without unit, the same as in the last example if we specialize to the case m = p. In particular, Ke(R, 1) = Ke(Z, (p)). It is a result of Rim, which we will discuss later on, that the map R -+ S induces an isomorphism on Ke. In particular, &(R) 2 C(S), the class group of the cyclotomic field. This is known to be non-zero for primes p 2 23. (See Example 3.3.5(b) below.) The smallest group G for which Ko(ZG) is non-trivial is the quaternion group of order B-in this case, Ke(ZG) is of order 2 and an explicit generator is exhibited in Exercise 1.7.20(3) below. 1.5.11. Exercise. The excision theorem may be interpreted as saying that the split exact sequence 1.5.2 gives rise to a split exact sequence of Ke-groups, the first group of which is Ke(1). The same holds by definition in the case of the split exact sequence 1.5.8. Using ideas from the proof of the excision theorem, show that if O-,I-+R-,R/I+O is split exact (i.e., I is an ideal in a ring R, and there is a splitting home morphism R/I 4 R), then

o ---) &(I) -+ I%(R) -+ &(RII) * 0 is split exact.

6. An application: Swan’s Theorem and topological K-theory To many mathematicians, the term K-theory suggests not algebraic K theory but topological K-theory, an exceptional cohomology theory 01 compact Hausdorff spaces defined using vector bundles. The connectio: between the two comes from specializing what we have done to the cas where R is a ring of continuous functions. In this context, the Excisio

32

1. KO of Rings

when G is free abelian; this case will be treated in Chapter 3. For finite groups, Ke(ZG) is often non-trivial and contains interesting arithmetic information. Consider the simplest example, when G is cyclic of prime order p, say with generator t. Then R = ZG may be identified with ?I&]/(@’ - 1). If [ = e2ri/P, a primitive pth root of unity, and if S = Z[

Ex,E4E a n d FxE+E which restrict to vector addition and scalar multiplication on each fiber. Such bundles E 3 X make up a category, in which the morphisms are commutative diagrams f E E’ P

1

P'

1

X--- X for which the map E f E’ is linear on each fiber. For any X and any n E N, the category always includes the trivial Fvector bundle of rank n, which is X x Fn -% X, where ~1 is projection on the first factor and the vector bundle structure is the obvious one coming from the vector space structure on the second factor. The category has a binary operation called the Whitney sum, denoted $. By definition, if E 5 X and E ’ L X are F-vector bundles over X, their Whitney sum is defined by E CB E’ = {(z, z’) : z E E, x’ E E’, p(z) = p/(x’)}, with the obvious map to X. For most purposes we want a more restrictive definition. A (locally trivial) F-vector bundle is a F-vector bundle in the above sense with the additional property that for each x E X, there is a neighborhood U of x in X and an isomorphism (in the category of F-vector bundles) from p-l(q p'p-y U to a trivial bundle of some rank over U. The rank of such a bundle is then a continuous function X + N defined by rank,(E) = dimp-l(x). If X is connected, the rank must be constant.

34

1. Ko of Rings

1.6.2. Definition. If X is a compact Hausdorff space, let Vectp(X) denote the monoid of isomorphism classes (in the category of IF-vector bundles) of locally trivial F-vector bundles over X, with an addition operation induced by the Whitney sum. The O-element of this monoid is the trivial bundle of rank 0. The topological K-theory of X is defined by K,(X) = G(Vectp(X)). S ometimes this is denoted simply K(X) or “0” stand respecKU(X) if F = C, KO(X) if F = Iw. (The “U” tively for “unitary” and “orthogonal” after the names of isometric linear transformations.) We will often suppress mention of F when it is understood from context. If X is connected, the reduced topological K-theory is K:(X) = k e r ( r a n k K”(X) actually is a contravariant functor from the category of compact Hausdorff spaces (and continuous maps) to the category of abelian groups. This follows from the fact that vector bundles pull back under continuous maps. If X f Y is continuous and E -% Y is a vector bundle over Y, we define f*(Y) to be the fiber product

{(x, e) : x E X, e E E, f(x) = p(e)), with the obvious map to X. The pull-back clearly induces a monoid home morphism f’ : Vectp(Y) -+ Vectp(X) and thus a map K’(Y) -+ K’(X). We’re now ready for the connection between vector bundles and projective modules that explains the connection between topological and algebraic K-theory. 1.6.3. Theorem [Swan2]. Let F = Iw or Cc, let X be a compact Hausdorff space, and let R = C*(X) be the ring of continuous F-valued continuous functions on X (with pointwise addition and multiplication). If E 5 X is a (locally trivial) F-vector bundle over X, Jet r(X, E) = {s : X -+ E continuous 1 p 0 s = idx} be the set of continuous sections of p. Observe that this is naturally an Rmodule. Then I’(X, E) is finitely generated and projective over R, and every finitely generated projective module over R arises (up to isomorphism) from this construction. The map E y-1 I’(X, E) induces an isomorphism of categories from the category of (locally trivial) vector bundles over X to the category of finitely generated projective R-modules. It also induces an isomorphism K’(X) -+ Kc(R). Proof. Let E 3 X be a (locally trivial) F-vector bundle over X and let I’(X, E) be its R-module of sections. For each x E X, there is an open neighborhood U over which E looks like a trivial bundle U x IF” for some n. The n constant functions ej : U --t JR’” determined by the standard basis vectors of IF” clearly generate the sections of this trivial bundle as a module over the continuous functions. Since X is compact, we can cover X by finitely many such open sets Vi and choose a partition of unity (fi)

6. An application: Topological K-theory

35

subordinate to the covering. (Thus 0 5 _fi 5 1, fi is supported in Vi, and Cfi E 1.) MuItip1ying the ej corresponding to Vi by fi, we get sections eij supported in Ui which clearly extend to all of X by taking them = 0 off Vi, and by construction, the eij generate I’(X, E) as an R-module. Hence I’(X, E) is finitely generated. Next we show that I’(X, E) is projective. Choose generators sj, 1 5 j 5 Ic, for l?(X, E) as an R-module. (These may or may not be the ones we just constructed above.) Consider the trivial bundle X x Fk z X and construct a morphism ‘p : X x IF” + E by k

(z,

w,... ,wk) I--+

cWjSj(X). j=l

Since the sj( z )span p-‘(z) for each 2, this vector bundle morphism is surjective on each fiber. Define a subbundle of the trivial bundle by E’ = ker cp , i.e., by EL = ker cpz. This is also locally trivial since one can check that it is trivial over any open set where E is trivial. We claim now that E @I E’ g X x Fk, which will show that

ryx, E) 63 r-(x, E') E ryx,

X x

lFk) 2 R”,

hence that r(X, E) is a projective module over R. The easiest way to do this is by introducing hermitian metrics, i.e., inner products. A hermitian metric on E is a continuous map (, ):ExxE+F which restricts to a positive-definite inner product on each fiber of E (bilinear if IF = R, sesquilinear if IF = Cc). Such metrics clearly exist since they exist on trivial bundles (use the standard inner product on F”) and can be patched together using a partition of unity. Therefore we may choose such a metric on E and the metric on X x Fk coming from the standard inner product on IF’“. With respect to these metrics, cp has an adjoint cp* satisfying the usual relation

(VJ, 4 = (v, cp*4. Since cp is surjective on each fiber, cp* will be injective on each fiber, with image the orthogonal complement of E = ker cp. So ‘p* gives an isomorphism of vector bundles from E to E”, showing that E 63 E’ 2 X x Fk, as desired. Now we have to show that every finitely generated projective module over R corresponds to a vector bundle. Suppose P is such a module and P@Q = R” s C(X, F”) . Then we may view P as a collection of functions X --+ IF” and let E = ((2, ~1,. . . , v,) E X x IF” : 3s E P with s(z) = (VI,. . . , w,)}.

36

1. KO of Rings

Define p : E + X using projection onto the first factor. It is now quite easy to see that E 5 X is a vector bundle. Vector addition and scalar multiplication just come from vector addition and scalar multiplication in Fn. (These operations map E into itself since P is an R-module.) We need only check the local triviality. Given z E X, choose elements el,. . . , eT E P such that el(z), . . , er(z) are a basis for the subspace E, = p-‘( z) of F”. Recall these are vector-valued functions; write ei = (ei, . . . , ek). Then since e1(2),..., er (x) are linearly independent, we can choose 1 I jl < . . . < j, < n such that (1.6.4)

e = det

is non-zero at x. We may choose similar elements f’, . . . , f n--T E Q such that f1 (x), . . . , fnpT(x) are a basis for the image of Q in F” at x. (The dimensions are complementary since P $ Q = R” ?Z C(X, IF”) .) From the f” we may construct an (n - r) x (rz - r) determinant f, similar to (1.6.4), which is non-zero at x. Since e and f are continuous, there is some neighborhood U of x in which both e # 0 and f # 0. For y E U, e ’( y),..., e’(y) are linearly independent and generate a rank-r free submodule of P. Similarly, f1 (y), . . . , f”-‘( y) are linearly independent and generate a rank-(n - T) free submodule of Q. By dimension counting, these must exhaust P and Q, so both P and Q are trivial over U. The statement about an equivalence of categories is now easy to check. 0 Theorem 1.6.3 suggests that we should extend the definition of K” to the category of locally compact spaces and proper maps (maps that extend continuously to the one-point compactification) by letting K”(Y) = Ko(C:(Y)), wh ere C:(Y) is the ring of functions vanishing at infinity on Y and we are using K-theory for rings without unit, as in Definition 1.5.7. The resulting theory is called K-theory with compact supports. See Exercise 1.6.14 below for a more geometric definition. 1.6.5. Proposition. If X is a compact Hausdorff space and A is a closed subspace, there is (for F = either Cc or Iw) an exact sequence induced by the inclusion A -+ X: K”( X \ A) -+ K’( X) + K’( A). Proof. Let R = CF(X), and let I be the closed ideal of functions vanishing on A, which as a ring without unit is isomorphic to C,“(X \ A), the functions vanishing at infinity on the locally compact space X \ A. By the Tietze Extension Theorem, every continuous function on A is the restriction of a continuous function on X, hence R/I may be identified with CF(A), with the quotient map R + R/I identified with restriction of functions. The result now follows immediately from Theorem 1.6.3 and Theorem 1.5.9. 0

6. An application: Topological K-theory

37

Proposition 1.6.5 shows in effect that K” satisfies two of the EilenbergSteenrod axioms for a cohomology theory: exact sequences and excision. It also satisfies the other key axiom, homotopy invariance, and we prove this next by using special properties of Banach algebras. Recall that a Banach algebra A is an algebra over R or Cc which also has the structure of a Banach space, such that for any a, b E A, ljabll < Ilallllbll. The principal examples for our purposes are A& (C” (X)) , X a compact Hausdorff space, or M,(C:(Y)), Y a locally compact Hausdorff space. The latter does not have a unit. 1.6.6. Lemma. Let A be a (real or complex) Banach algebra with unit and let x E A with 111 - 211 < 1. Then for each a E R there is an element xa in A with the usual properties (x1 = x, x0 = 1, xa . xp = xa+o). In particular, x is invertible in A. Proof. Define x” by the usual binomial power series for (1 + (x - l))a. The norm of the n-th term in the series is bounded by the corresponding term in the series for (l+lls-lll)a, which converges absolutely. Since A is a Banach space, the series for xa therefore converges absolutely. The relation $a . ,P - xa+@ follows as usual from multiplication of the series. Cl

1.6.7. Lemma. Let A be a Banach algebra and let p, q be two idempotents in A with ((p - q(I < 1. Then the projective A-modules Ap and Aq are isomorphic. Proof. Observe that pAp and qAq are Banach algebras with unit elements p and q, respectively. Since Ilp - pqpll < 1 and IJq - qpqll < 1, x = @qp)-i makes sense in pAp and qpq is invertible in qAq, both by Lemma 1.6.6. Thus there is an x E pAp commuting with pqp with x2(pqp) = p and of course with x = xp = px. Observe then that

(1.6.8)

(4 (W) = XPqPx = x2 (pqpp> = P,

that (1.6.9)

PC4 = ?I = (ed%

c?(v) = clx = (VIP>

and that (1.6.10)

(qx2d(4Pd = QX2clPcl = Qx2(PqP)q = qpq.

The equation (1.6.10) says (qx) (xq) is a left unit for qpq in qAq. But since qpq is invertible in qAq, (qx)(xq) must be equal to the unit element of qAq, which is q. The equations (1.6.8) and (1.6.9), together with this fact, imply that right multiplication by qx gives an isomorphism from Aq onto Ap, whose inverse is right multiplication by xq (compare the calculation in Lemma 1.2.1). Cl

38

1. Ko of Rings

1.6.11. Corollary (Homotopy invariance of topological K-theory). Let A and B be Banach algebras and let vt : A + B, 0 5 t 5 1, be a homotopy of homomorphisms from A to B. (This means exactly that there is a homomorphism ‘p : A + C( [0, 11, B) which when composed with evaluation at t gives cpt.) Then cpo and cp1 induce the same map on K-theory Ko(A) -+ &(B). Proof. If necessary, adjoin units to A and B and extend vt to a homotopy of unital homomorphisms of unital algebras A+ -+ B+. Since Ko(A) -+ Ko(A+) and Ko(B) - Ko(B+), this reduces us to the unital case. For simplicity, we therefore assume without loss of generality that A, B, and the homomorphisms are unital. For any p E Idem( p lies in M,(A) for some n, so we may replace A and B by M,(A) and M,(B), respectively. (These are still unital Banach algebras, and pt extends naturally to M,(A) just by application of the homomorphism to each matrix entry separately.) Then q(p) is a continuous path of idempotents in B. We may partition the interval [0, l] into subintervals such that Ilqt(p) - cps(p)ll < 1 for t, s in the same subinterval. By Lemma 1.6.7, the class of cpt(p) remains constant in each subinterval, hence remains constant in the whole interval. So cpo and cp1 induce the same map Idem + Idem and hence the same maponK0. q

1.6.12. Corollary. The functors X -+ VectF(X) and X or) K”(X) are homotopy-invariant functors from the category of compact Hausdorff top& logical spaces to the category of abelian monoids and the category of abelian groups, respectively. In particular, if X is contractible, all vector bundles over X are trivial, and PO(X) = 0. Proof. Specialize to the case of Banach algebras of the form M, ( CF (X)) . Since homotopic idempotents are equivalent, we deduce that the map from X to isomorphism classes of direct summands in a trivial bundle of rank n over X is a homotopy functor. The rest of the statements follow from this. Cl

1.6.13. Example. Corollary 1.6.12 shows that the classification of vector bundles, and hence the calculation of K’(X), are homotopy-theoretic in nature. Consider for instance the case where X = 5’“. This is a union of two contractible hemispheres joined along the equator S”-l. (If n = 0, the hemispheres are single points and the “equator” is the empty set.) Thus any rank-r bundle over X is trivial over the hemispheres and determined by the homotopy class of the “gluing data” along Y = P-l, which gives an isomorphism between the two trivializations of the bundle coming from the two hemispheres. Now an isomorphism between two trivial bundles Y x F’ 3 Y is just given by a continuous map Y -+ GL(r, F). So isomorphism classes of rank-r F-vector bundles over S” are in one-to-one correspondence with homotopy classes of maps P-l ---f GL(r, IF). Furthermore, by polar decomposition, any matrix in GL(r, F) can be written uniquely in the form up, where u is unitary if F = Cc, orthogonal if F = R, and p is positive-definite self-adjoint. The positive-definite self-adjoint ma-

6. An application: Topological K-theory

39

trices form a contractible space (since one can write any such matrix as eh with h hermitian and use the contraction given by eth, 0 5 t 5 l), so GL(r, Cc) has a deformation retraction to U(T) and GL(r, R) has a deformation retraction to O(T). Thus the isomorphism classes of rank-r IF-vector bundles over S” are given by 7r,_i(U(r)) if IF = Cc, 7r,-1(0(r)) if IF = R. The O-element of the homotopy group corresponds to the trivial bundle. Now we can make some computations. O(r) always has two components with identity component the rotation group SO(r), and U(T) is connected. Thus re(U(r)) = 0 and rre(O(r)) F Z/2, so %?‘(S’) = 0, E’(S’) Z Z/2. In low dimensions, one can check that O(1) = (1, -l}, SO(2) g S1, SO(3) E lRP3, SO(4) has S3 x S3 as a double cover, U(1) FZ S’, SU(2) cz S3. Thus, for instance, 0, r = 1, W(O(T)) =

z, T = 2, ( 212,

r 2 3,

so that Vectn( S2) is the monoid described in Exercise 1.1.7. One finds sim- 0 ilarly that rl(U(r)) = Z for all r, so that KU (S2) 2 Z. The calculations of K?j”(Sn) and of E’(F) for all n follow from the Bott Periodicity Theorem, which says that the answer only depends on the value of n mod 8 in the real case or the value of n mod 2 in the complex case. One obtains 0 , r$O,1,2,4 mod8, Z, r=O,4 mod8,

E’(F) =

212,

rrl,2 mod8,

1.6.14. Exercise. Give another description of K-theory with compact supports for a locally compact Hausdorff space Y in which K’(Y) is a set of equivalence classes of triples (Ee, El, ‘p), where Es and Ei are (locally trivial) vector bundles over Y and cp is a morphism of vector bundles Ee + El which is an isomorphism outside of a compact set, and with relations

(b)

[Eo, El, p] = [Eo, El, cp’]

(cl

[Eo, El, ‘p] = 0

if cp = ‘p’ outside of a compact set, if cp is an isomorphism.

Impose the necessary equivalence relation to get an isomorphism with our old description of K”. Hint: when Y is actually compact, condition (b) says that one can forget the cp altogether. In this case, the isomorphism of this description of K” with the usual one is given by [Eo, El,

‘PI ,-+ PO] - PII.

1. KO of Rings

40

1.6.15. Exercise. Show that if one defines K-j(X) = K”(X x ll@) (using K-theory with compact supports) that the short exact sequence of Proposition 1.6.5 can be extended to a long exact sequence . . . --f K-j(X \ A) --+ K-j(X) + K+(A) + K-j+‘( X \ A) -i . . . . Hint: the problem is construct the boundary map KO(AxR) + KO(X\A). This can be done by letting Y be the space (A x (0, 11) U X, with (a, 1) identified with a E X for a E A. (Y is the “open mapping cone” of the inclusion A L) X.) One gets from Proposition 1.6.5 exact sequences K”( A x (0, 1)) -+ K’( Y) + K’( X)

and K”( X \ A) + K’( Y) + K”( A x (0, 11).

Show using homotopy-invariance and excision that K”( A x (0, 11) vanishes and that K”( X \ A) + K”( Y) is an isomorphism. Then splice these exact sequences together with the sequence K”( X \ A) -+ K’( X) + K’( A). 1.6.16. Exercise (The Karoubi Density Theorem [Karoubi, 11.6. 151). Let A and A be (unital) Banach algebras over Cc, and let L : A + A be a continuous injection of A into A as a dense subalgebra. Extend L

to matrices in the usual way, by applying it to each entry of the matrix. Assume that for all n, if z E M,(d) andL z( is) invertible in M,(A), then z is invertible in M,(d). (1) Show that L induces an isomorphism Ko(d) + Ko(A). Hint for the surjectivity: if e is an idempotent in h&(A), then e can be approximated in the topology of A by an element z of M,(d). Show that the spectrum of x in M,(d) coincides with its spectrum in M,(A), and thus that x has spectrum close to (0, 1). Deduce that the Banach subalgebra of M,(d) generated by x contains an idempotent f with I c1 ose to e, by justifying the definition

f=& Jy&J

(2)

where I is a contour in the complex plane encircling the part of the spectrum of x close to 1, and excluding the part of the spectrum of x close to 0. Then use Lemma 1.6.7. Show that the two hypotheses are satisfied if A is the algebra of continuous complex-valued functions on a compact subset X of R” (equipped with the sup norm I] I]), and if A is the algebra of continuously differentiable functions on X, equipped with the norm

Ilf IId = Ilf I + llvf 11. Deduce that “every vector bundle over X has a differentiable structure.”

7. Another application: the Wall finiteness obstruction

41

7. Anot her application: Euler characteristics and the Wall finiteness obstruction In this final section of Chapter 1, we discuss the algebraic background of most of those applications of Ke to topology that do not involve topological K-theory. While what we will be doing here is pure algebra, it is worth saying a bit about the topological motivation to explain what is going on. If X is a path-connected, locally l-connected topological space with fundamental group G and R = ZG, we can manufacture from X its singular chain complex with local coefficients S.(X). This is a chain complex of free R-modules which is the same thing as the usual singular chain complex of the universal cover X of X, together with the R-module structure coming from the action of G on X by covering transformations. Furthermore, the chain homotopy equivalence class of the chain complex S.(X) only depends on the homotopy equivalence class of the space X. The chain complex S.(X) is quite large in general; for most spaces of interest, the R-modules in it are not even countably generated. However, if X is a finite CW-complex, then S.(X) is chain homotopy equivalent to the cellular chain complex with local coefficients C.(X), a chain complex of free R-modules with only finitely many non-zero chain groups and with each of these chain groups finitely generated. Thus an obvious necessary condition for a space X to be homotopy-equivalent to a finite CW-complex is for S.(X) to be chain-homotopy-equivalent to a finitely generated complex of free R-modules. Under some circumstances, it is easy to check not this condition but something weaker, called finite domination. The space X is finitely dominated if up to homotopy it is a retract of a finite CW-complex; in other words, if there is a finite CW-complex Y and there are maps f : X --+ Y, g : Y + X with g o f Y idx. An important question is then whether this implies that X is homotopy-equivalent to some (other) finite CWcomplex. (It is not hard to show that X is homotopy-equivalent to some CW-complex (see [Varadarajan, Theorem 3.91 or [Spanier, Ch. 7, Exercise G6]), but this complex is not necessarily finite.) This question was answered by C. T. C. Wall in an important series of papers. Wall showed that if X is finitely dominated, then S.(X) is chain-homotopy-equivalent to a finitely generated complex of projective R-modules. The Wall finiteness obstruction of X is then the “Euler characteristic” of this complex in the group l&,(R). Though we will not show here that vanishing of the obstruction is sufficient for finiteness (for this see [Wall] or [Varadarajan]), it will be clear that it is necessary. The Wall obstruction occurs in many problems in geometric topology, such as the question studied by Siebenmann of when a non-compact manifold is homeomorphic to the interior of a compact manifold with boundary. For this and other geometric problems related to the Wall obstruction, see [Weinberger, Ch. 1, 31 and 941. We shall now provide an abstract treatment of the Wall finiteness obstruction for chain complexes of R-modules, as an outgrowth of the classical theory of the Euler-Poincare characteristic for topological spaces. Since we

42

1. Ko of Rings

don’t assume the reader is very familiar with homological algebra, we begin with a review of some classical notions and facts. The reader who has had a course in homological algebra or homology theory can probably skip ahead to 1.7.9 after reviewing the statements of Theorems 1.7.4 and 1.7.7. 1.7.1. Definition. Let R be a ring (with unit). A chain complex of R-modules is a pair (C., d), where C. is a Z-graded R-module and d is an R-module homomorphism C --f C of degree -1 such that d2 = 0. (In other words, d is defined by maps d, : C, ---f C,_, such that d,_l o d, = 0.) Recall that the homology of such a chain complex is H(C) = ker d/ im d; more precisely, H, = ker d,/ im d,+l. Elements of ker d are called cycles and elements of imd are called boundaries. The chain complex is called acyclic if H(C) = 0, i.e., if the sequence &+I 6-l ... G 5 C,_l - . . . is exact. 1.7.2. Definition. If (C., d), (CA, d’) are chain complexes of R-modules, a chain map between them is an R-module homomorphism cp : C -t C’ of degree 0 intertwining d and d’, i.e., is given by maps (P* : C, + Ck such that dk o pn = pn_l o d,. It is immediate that such a cp induces maps on homology cpt : H,,(C) --+ H, (Cl). If ‘p : C -+ C’ and 7c, : C --) C’ axe chain maps, a chain homotopy between them is an R-module homomorphism s : C --) C’ of degree +l such that (1.7.3)

sod+d’o s=cp-$.

Chain homotopy is an equivalence relation on chain maps. We write ‘p N_ 1c, if there is a chain homotopy between them. A chain homotopy from idc to 0 is called a chain contraction, and if such a homotopy exists, C. is called chain-contractible. Note that (1.7.3)lm ’ pl ies that ‘p* = & on homology. Indeed if dx = 0, then cp(z) - +(z) = s o d(z) + d’ o s(z) = d’( s(z)), so that p(x) and $(z) lie in the same homology class. Thus if a chain complex is chain-contractible, it is acyclic. The converse is false without additional conditions. If there exist chain maps cp : C ---f C’ and I/J : C’ + C such that $0 cp N idc and cpo+ N idcl , then we say C and C’ are chain-homotopyequivalent. This of course implies by our previous remark that cp* is an isomorphism on homology with inverse $J*. 1.7.4. Proposition. If (C., d) is an acyclic chain complex of projective R-modules and C. is bounded below, i.e., Cj = 0 for j sufficiently small, then C, is chain-contractible. Proof Without loss of generality assume Cj = 0 for j < 0. (Otherwise reindex.) We construct a contraction s, : C, -+ C,+l by induction on n to satisfy the needed condition (*n)

s,_l o d, + d,+l o s, = idc,.

7. Another application: the Wall finiteness obstruction

43

At the same time, we also show by induction that kerd, is a direct summand in C,. To begin the induction, set sj = 0 for j < 0 and note that by the assumptions that Ho(C) = 0 and C-i = 0, dl : Ci + Cc must be surjective. Since Cc is projective, dl must have a right inverse SO, so (*e) holds. Furthermore, im dl = ker do = CO is projective. For the inductive step, assume we’ve constructed sj for j < n to satisfy (*j) and we know ker dj = imdj+i is a direct summand in Cj for j < n, hence projective by Theorem 1.1.2. We shall construct s, to satisfy (*n). By inductive assumption, C,_i = (im d,)@Q,_l for some projective Q+i. On imd, = kerd,_i, s,_2 0 d,_l = 0, so d, o s,_l is the identity. Thus, by (*+I), s,-1 is a right inverse for d, : C, + imd, C: C,_i.

Therefore s,_i o d, is an idempotent endomorphism of C, with image Qn complementary to ker d,, and ker d, = im d,+l is R-projective. Since dn+l

. G,l +

imd,+i = ker d,

is surjective, it has a right inverse s,. Extend s, to all of C, by making it 0 on Qn. Then (*n) is satisfied and we’ve completed the inductive step. The Proposition now follows by induction. 0 1.7.5. Definition. Suppose ‘p : (C., d) + (CL, d’) is a chain map between chain complexes of R-modules. Its mapping cone is (C:l, d”) , where Cf = C,_i @ Ci (note the degree shift in the first summand!) and d;(c, c’) = (-d+lc, v(c) + d;(c’)).

This is a chain complex since d;_l o d;(c, c’) = (dj-2 o d+lc, cp(-dj-lc) + d[i_l (v(c) + d;(c’))) = (0, -cp o dj-l(c) + d;_l 0 v(c) + 0) = (0, 0).

1.7.6. Theorem (Fundamental Theorem of Homological Algebra). Suppose

0 --t

(CL,

d’) 5 (C., d) 2 (C:‘, d”) + 0

is a short exact sequence of chain complexes. (This means a and /3 are chain maps and the sequence of R-modules

is exact for each j.) Then there is an induced long exact sequence of homology modules

--dzj(c’) =Hj(c) % Hj(C”)

3 Hj-l(C’)

+....

44

1. KO of Rings

Proof. This is the quintessential LLdiagram chase.” First we go through the definition of Hj(C”) 2 Hj-l(C’); then we go through the proof of exactness. Let [x”] be a class in Hj(C”) represented by x” E Cy with d”x” = 0. Since p is surjective, x” = P(x) with x E Cj. Since d”o/3(x) = 0 and p is a chain map, ,0 o d(x) = 0, i.e., d(x) E kerp = imcu. Hence d(x) = a(~‘) for some x’ E Cj_1. We claim d’( x’) = 0, so that x’ is a “cycle,” i.e., represents a class in Hj-1 (C’) . Indeed, since (Y is a chain map, cr: o d/(x’) = do ‘X(X’) = d2(x) = 0. But a was injective, so d/(x’) = 0. Now let a[~“] = [&I. We 1eave to the reader the simple argument that shows this is independent of the choice of x” within its homology class and independent of the choice of the lift x of 2”. We proceed now to the proof of exactness. The construction of a[~“] above gives cr*(a[x”]) = [a(~‘)] = [0], and also shows that if [x”] = p*[x] for some [x] E Hj(C), then a[~“] = 0 (since d(x) = 0). Also, /% 0 Q* = 0 since 0 o (Y = 0. So the image of each map in our sequence is contained in the kernel of the next one. For the reverse containments, suppose for instance that x E Cj, d(x) = 0, and fl*[x] = 0 in Hj(C”). Then P(x) = d”(y”) for some 9” E Cy+i. Since /I is surjective, we may choose y E Cj+i with p(y) = y”. Since d”o&) = Pod(y) = P( x ) , x-d(y) E kerp = ima, and [x] E imo,. Thus ker/3* C imcr,. N e x t , s u p p o s e x E C$‘, d ”( x) = 0, and a[~“] = 0 in Hj_l(C’). By the description of d above, this means x” = p(x) with d(x) = cr(x’) and x’ = d’(y’), y ’ E Ci. T h e n d o a(~‘) = (Y o d’(y’) = cr(x’) = d ( x ) , s o x - (I E kerd. Since also P(x - cr(y’)) = P(x) = x”, this shows [x”] = ,&[x - cr(y’)]. Hence kera c imp*. Finally, suppose X’ E Cj-1, d’( x’) = 0, and CX*[X’] = 0 in Hj_l(C). Then CX(Z’) = d(x) for some 2 E Cj. Let x” = P(x). Then d//(x”) = P 0 d(x) = Doa = 0, so x” defines a class [zE”] in Hj(C”). From the description of d, a[~“] = [x’], so ker (Y* C: im a. This completes the proof of exactness. 0 1.7.7. Theorem. A chain map between chain complexes of R-modules is a chain homotopy equivalence if and only if its mapping cone is contractible. If the complexes are

b e l o w

a n do f

R - m o d u

then it is a homotopy equivalence if and only if the mapping cone is acyclic, ’ d) L e tcp (C., --+ be a chain map and let (Ct , d”) be its mapping cone. First observe that there is a short exact sequence of chain complexes 0 + (CL, d ’) --f (C:, d ”) + (C.-l, -d) + 0.

The maps here are the obvious ones: we map C; to Cy = Cj_i @ C(i by c’ H (0, c’), and we project Cy onto the first summand Cj-1. The fact that these maps commute with the boundary maps is obvious from Definition 1.7.5. Since changing the sign of d doesn’t change the homology of C, we obtain from Theorem 1.7.6 an exact sequence (1.7.8)

... -+ HJC”) 4 H,_l(C) 5 H,_l(C’) + H,_l(C”) + ... .

7. Another application: the Wall finiteness obstruction

45

Here it is easy to check from the definition of d that the map H,_I(C) ---) Hn_l(C’) is just cp+. Thus cp* alentbto 19” e ithe n gmapping a c y c lcone ic. Furthermore, if C and C’ are bounded below and consist of projective modules, then the same is true of C”. Hence, by Proposition 1.7.4, the mapping cone in this case is acyclic if and only if it is contractible. It remains to show that cp is a homotopy equivalence if and only if C” is contractible. Suppose s” : C” --+ C” is a chain contraction. Then we define s : C 4 C, s’ : C’ + C’, and 1c, : C’ --) C b y s”(C, 0) = (s(c), *. .), s”(0, c’) = ($!l(c’), -s’(d)). Since d”s” + s”d” = idp, we have (c, 0) = (-do s(c), se.) + s”( -dc, p(c)) = (-do s(c) + + o v(c) - s o d(c), . . .) , (0, c’) = d”($(c’), -s’( c))) + ~“(0, d ’( c’) ) = (-do $(c’), cp o $(c’) - d’ o s’(c’)) + ($ o d’(d), -s’ o d'(c')) , which says - do $(c’) + 11, o d’( c’) = 0 Vc’

($I is a chain map);

c=-dos(c)++ocp(c)- s o d ( c ) Vc ($opc~“idc); S c’ = cp o $(c’) - d’ o s’(c’) - s’ o d’(c’) Vc’ (‘p o $ 5 idct).

( In the other direction, suppose cp is a homotopy equivalence with homotopy inverse $J, and suppose one has homotopies s from $J 0 ‘p to idc and s’ from cp o $ to idcj. Let s”(C, c’) = (s(c) + $(c’) + $0 s’ 0 cp(c)ll, 0 cp 0 s(c), -s’(c’) + s’ 0 $9 0 s(c) - (s’)2 0 cp(c)) . We will check that one obtains a chain contraction of C”. Note that (d”s” + s”d”) (c, c’) = d” (s(c) + T+!J(c’) + II, o s’ o p(c) - $0 ‘P o s(c), -s’(c’) + s’ 0 p o s(c) - (s/)2 o p(c)) + s” (-d(c), p(c) + d’(d)) = (-do s(c) - do $(c’) - do v+b o s’ o p(c) + do $ o cp o s(c), p o s(c) + cp o I@‘) + $0 o $ o s’ o p(c) - cp o 1c, o ‘p o s(c) -d’ o s’(c’) + d’ o s’ o cp o s(c) - d’ o (s’)~ o v(c)) + (-s o d(c) + $J o p(c) + $0 d’(c’) + $J o s’ o 4-d(c)) + 1c, o ‘p o s(d(c)), - s’ o v(c) - s’ o d’(c’) +s’ o cp o s(-d(c)) - (s’)~ o p(-d(c))) .

46

1 . KoofRings

The first coordinate (after some regrouping) is

$(c+’) $J 0 d’(c’)l

= [-d 0 s(c) - s 0 d(c)] +

=

+[-doIjlos’ocp(c)-II,os’ocpod(c)] + [d 0 7+h 0 cp 0 s(c) + 1c, 0 cp 0 s O 441 + + O cp(4 - (dos+sod)(c) + (11,ocposod+~ooodoss)(c) - 6 o (d’ o s’ + s’ o d’) o p(c) + + o v(c)

= c + 1c, o cp o ($ o cp - idc)(c) - $ o (cp o 1cI - idc’) o ~(4 = c. The second coordinate (after some regrouping) is = 19 o $(c’) - d’ o s’(c’) - s’ o d’(c’)] + [p’ o o s (sc’) + d o cp o s ( c-p) o 11, o cp o ’ o v(c) - ’ o (s’)’ o v(c) - s’ c’

cp o s o d(c) +

(idct - cp o+

cp o$

s(c)]

’ o p(c)]

’) ~ cp o

’ o s’) o cp

(‘p o $I - idc, - d’ o s’) + - ’ ocp

’ o v(c)

(s’)~ o cp o d(c)

= c’ - s’ o d’ o cp o s(c) + s’ o d’ o s’ o p(c) - s’ o cp o s o d(c) + (s’)~ o cp o d(c) =c’-s’ocpo(dos+sod)(c) + s’ o (d’ o s’ + s’ o d’) o p(c) = c’ - s’ o cp o (T+!J o cp - idc)(c) + s’ o (‘p o 1c, - idct) o v(c) =c.I This confirms that s” is a chain contraction of C”.

0

Now we’re ready to introduce the connection with Ko(R). 1.7.9. Definition. A chain complex (C., d) of R-modules is called bounded if the modules Cj are non-zero for only finitely many j, and is called of finite type if it is bounded and all the Cj are finitely generated. (The connection with topology is that the cellular chain complex of a finite CW-complex is of finite type, and the cellular chain complex of a finitedimensional CW-complex is bounded (with non-zero chain groups only in the dimensions of the cells of the complex) .)

7. Another application: the Wall finiteness obstruction

47

If (C., d) is a chain complex of finite type of projective R-modules, we define its Euler characteristic by x(C) = 2 (-l)j[Cj] ( i n K c ( R ) ) . j=-_oo Note that this is really a finite sum, and that d is not used in the definition of x(C). Also define g(C) to be the image of x(C) in kc(R). Proposition (“E uler-PoincarB Principle”) . The Euler 1.7.10. characteristic is additive on short exact sequences of complexes of finite type. In other words, if

is a short exact sequence of chain complexes of finite type of projective R-modules, then x(C”) = x(C’) +x(C). tithermore, if (CL, d) is a chain complex of finite type of projective R-modules, and if all its homology modules are projective, then

Proof. Since any short exact sequence of projective modules splits, if

short exact sequence of chain complexes of finite type of projective R-modules, then Cy G+ Ci @ Cj, hence [CT] = [C$] + [Cj] and the formula x(C”) = x(e) + x(C) follows upon taking the alternating sum over j. Next, suppose (C., d) is a chain complex of finite type of projective R-modules and all the homology modules Hj(C) are R-projective. Let Zj = ker(dj : Cj + Cj-I), Bj = im(dj+r : Cj+r -+ Cj). We have short exact sequences

is a

(*)

0 + Zj+l

(**I

0

--) Cj+l

-+Bj

* Bj + Zj + Hj

+O,

+ 0.

Since Hj is assumed projective, (**) splits, and Zj E Bj @ Hj. Since the complex is assumed to be of finite type, we may assume (after reindexing) that Cj = 0 for j < 0, in which case Cc = Zc is projective; hence, since Zs G! B0 @ HO, BO is projective. Thus Ci 3 BO must split and so Ci % BO @I Z1. This implies 21 is projective, and since Zi g BI @ HI,

48

1. Ko of Rings

Bi is projective. Continuing by induction, all the Bj and Zj are projective and all of the above short exact sequences split. Thus we obtain (from (*)),

[zj+iI + [Rj] = [Cj+i] [Bj] + [Hj] = [Zj]

(from (**)).

Substituting in the definition of x(C), we obtain

x(C) = g (-VPjl j=-cc

=

2 (-l>j([Zj] + [Bj-I]) 2 (-l>j([f$] + [Bj] + [Bj-11)

j=-00

=

j=-00

= 2 (-l)j[Hj] j=-m

+ 2 (-l)j[Bj] 2 (-l)j[Bj] j=-00 j=-m

= 5 (-l)j[Hj]. j=-m

Cl

1.7.11. Corollary. The Euler characteristic is well defined on chain complexes of projective R-modules which are homotopy-equivalent to complexes of finite type of projective R-modules, and is constant on homotopy equivalence classes. It is also additive on short exact sequences of such chain complexes. Proof. Suppose (C., d) is a chain complex of projective R-modules which is homotopy-equivalent to a chain complex of finite type (C,l, dl) of pro’). jective R-modules. We define x(C) = x(C Of course, to know that this makes sense, we need to check that it is independent of the choice o f Cl. If (C.2, d2) is another possible choice, then C1 and C2 are each homotopy-equivalent to C, hence are homotopy-equivalent to each other. Let ‘p : Cl + C2 be a homotopy equivalence between them and let C3 be its mapping cone. Since Cl and C2 are of finite type and consist of projective R-modules, the same is true of C3. Furthermore, from the short exact sequence 0 -4 (C.“) --+ (C.“) --$ (Ci_,) -+ 0 and Proposition 1.7.10, we obtain that

x(C3> = xv”> - xv’>. But C3 is acyclic by Theorem 1.7.7, so its homology modules are 0 and hence x(C3) = 0 by Proposition 1.7.10 again. Thus x(C’) = x(C2),

7. Another application: the Wall finiteness obstruction

49

as required. The same calculation shows that homotopy-equivalent chain complexes have the same Euler characteristic. Finally, additivity on short exact sequences also follows immediately from Proposition 1.7.10 and the fact that short exact sequences of projective modules must split. 0 We’re now finally ready for Wall’s theorem. 1.7.12. Theorem [Wall]. Let (C., d) be a chain complex of projective R-modules which is homotopy-equivalent to a chain complex of finite type of projective R-modules. Then (C., d) is homotopy-equivalent to a chain complex of finite type of free R-modules if and only if g(C) = 0 in I&(R). Proof. Suppose (C., d) is homotopy-equivalent to (CL, d’) of finite type, with both complexes consisting of projective modules. By Corollary 1.7.11, x(C) = x(C’); h ence X(C) = g(C’). If C’ consists of finitely generated free modules, then clearly g(C’) = 0 so g(C) = 0. On the other hand, suppose g(C’) = 0. It will be enough to show C’ is homotopy-equivalent to a complex of finite type consisting of free Rmodules. Suppose Ci = 0 for j outside of an interval {k, k + 1,. . . , k + n}. Choose projective modules Qn, . . . , QO such that C(e+n @ Qn is free, C’k+7L_-1 $ Qn $ Qn_l is free, and in general such that CL+j @ Qj+l@ Qj is free for 0 5 j < n. If (T., dT ) is chain-contractible, then replacing (CL, d’) by (CL, d’) @I (T., 8) d oesn’t change its homotopy class. So let (Ti, 8’) be defined by 0,

Ti” = {

Qj,

i#k+j,k+j-1,

i=k+j,k+j-1,

with n. We first construct by induction on m, starting at 0 and continuing up to m = n, a complex of finite type (C$, d’)j n. To begin the induction, note that since Cj-1 = 0, Ho(C) = Co/ imdl. Choose a finite set of generators MY..., [zcT] for Ho(C) and representatives ~1,. . . ,z, E Co. Let CA be the free R-module on generators ~1, . . . , y,. and let cp~(yk) = zk. Since R is assumed Noetherian, the kernel BI, of the composite map

c:, = Co -+ Ho(C), being a submodule of the finitely generated module CA, is also finitely generated. Choose generators 21, . . . , zt for Bh and let C{ be free on generators wt. Define d’, so that di (wk) = zk. Then C: 3 CA is a chain com;:; wfth Ho(C’) = CA/B’o and cpo induces an isomorphism on HO. Since we want cpo to be the O-degree part of a chain map, we need to define cp1 so that G ‘p1 Cl

d:

1

c:,

dl

1

1po co

commutes. Since ‘pO(zk) goes to 0 in Ho(C), we can choose ‘Ilk E Cl, with dl(uk) = (PO(&). so we let pl(wk) = uk and the condition is satisfied. This completes the first step in the induction. For the inductive step, assume we’ve constructed a complex of finite type of free R-modules (Ci, dj) for j 5 m and a chain map cp : C’ + C which is an isomorphism on homology in degrees < m. Continuing as before, choose generators [xl], . . . , [x,] for H,(C) and representatives x1, . . . , x,. E zn c Gn. Replace the old Ck by its direct sum with the free R-module

7. Another application: the Wall finiteness obstruction

51

on generators yi, . . . , yr and let (~~(ylc) = xk. We keep qrn the same on the old CL. Similarly, we do not change d&, on the old C& and let dk(yk) = 0. Then we still have a chain complex and a chain map for j 5 m but now cp* is surjective on H,. As before, we choose CL,, finitely generated and free with d,+l : CL,, + C, mapping onto the kernel of the composite

and define vrn+i as above so that we have a chain map which now is an isomorphism in homology through degree m. We continue by induction until we’ve constructed a complex of finite type of free R-modules and a chain map cp : C’ + C which is an isomorphism on homology in degrees < n and a surjection in homology in degree n. Of course, since everything is zero in degrees > n, cp* is actually an isomorphism on homology in all degrees except n. Now consider the mapping cone (Ct , d”) of cp. This is a bounded complex of projective R-modules with non-zero chain modules only in degrees 0 through n + 1. By the exact sequence (1.7.8) (in our situation C and C’ are reversed), C” has only one non-zero homology module, in degree n + 1. Repeating the proof of Proposition 1.7.4, we can construct a chain contraction of C” through degree n, which shows that dE+, : CE+l + B” is split surjective and thus that Hn+l(C”) = ZE+l = ker dE+, is R-project&e and a direct summand in Ci+i Z CL. Hence we may replace CA by a projective complement to H,+l(C”) and thereby make C’ a complex of finite type of projective R-modules and cp* an isomorphism on homology, hence a chain-homotopy equivalence, by Theorem 1.7.7. 0 Remark. This proof demonstrates clearly the origin of Wall’s obstruction. At the last step of our induction, we can either make cp* into a homology isomorphism in degree n at the expense of making C, a possibly non-free projective module, or we can make Ck free and ‘p* an epimorphism on homolow in degree n, but in general we can’t take C,, free and at the same time make ‘p a homotopy equivalence. Now for some topological applications. Wall’s work on finiteness obstructions for chain complexes arose from the question of when a connected space X is homotopy-equivalent to a finite CW-complex. If Y is a finite connected CW-complex, Y is locally simply connected (so that covering space theory applies) and has a finitely presented fundamental group r. (The fundamental group of the l-skeleton of Y is a finitely generated free group surjecting onto 7r, and rr is obtained from this free group by adding in one relation for each 2-cell.) Thus we may form the universal covering Y of Y, which carries a free cellular action of 7r. The cellular chain complex of Y, while not of finite type over Z, may be viewed as a chain complex of finite type of free R-modules, where R = %, the integral group ring of T. Alternatively, we may think of this complex as the chain complex of Y with local coefficients. Thus if X is a space which is homotopy-equivalent to Y, it must also have fundamental group rr (finitely presented), and its singular chain complex with local coefficients S.(X), which is a complex of

52

1. Ko of Rings

free R-modules but is very far from being of finite type in general, must be chain-homotopy-equivalent to a complex of finite type of free R-modules. Theorem 1.7.12 now gives a necessary and sufficient condition for S.(X) to have this property. Call S.(X) finitely dominated if it is chainhomotopy-equivalent to a complex of finite type of projective R-modules. Theorem 1.7.12 says that a finitely dominated complex has a well-defined finiteness obstruction in &(R), and is chain-homotopy-equivalent to a complex of finite type of free R-modules if and only if this finiteness obstruction vanishes. If R = 221~ happens to be Noetherian, which is not the case for all finitely presented groups 7r, but is true say if 7r is a product of a finite group and a free abelian group (the group ring of a finite group is finitely generated as a Z-module, hence Noetherian, and the group ring of 7r x Zn is a Laurent polynomial ring in n variables over the group ring of 7r), one can apply Theorem 1.7.13 to see that an R-module chain complex C is finitely dominated if and only if it is homologically finite-dimensional and its homology groups are finitely generated. Wall actually went further than this; he showed that a connected space X with finitely presented fundamental group and the homotopy type of a CWcomplex is finitely dominated if and only if S.(X) is finitely dominated, and has the homotopy type of a finite CW-complex if and only if S.(X) is finitely dominated and has vanishing finiteness obstruction. The method of proof for the “if” directions is to inductively construct a sequence Y, (n 2 1) of finite CW- complexes by attaching cells, along with maps Y,, -+ X which are dominations (resp., homotopy equivalences) “through dimension n - 1.” The proof of Theorem 1.7.13 is an abstract version of this technique, in the case where R is Noetherian. In proving homotopy finiteness, the finiteness obstruction is precisely the obstruction to having this inductive process terminate after a finite number of steps. 1.7.14. Example. Let us illustrate a geometric application of Theorems 1.7.12 and 1.7.13. Suppose X” is a connected non-compact (topological or smooth) manifold and one wants to know whether X is homeomorphic to the interior of a manifold W” with boundary. Precise necessary and sufficient conditions were found by Siebenmann (provided one stays away from the problem dimensions 3 and 4 by assuming n < 2 or n 2 6) using surgery theory, but we have done enough now to at least give some interesting necessary conditions. If W” exists, then it must have the homotopy type of a finite CWcomplex, hence so must X, since the inclusion of X into W is a homotopyequivalence. Furthermore, for each component N”-l of aW, N must have a “collar” neighborhood in W homeomorphic to N x [0, l), so the corresponding “end” of X = W \ dW must be homeomorphic to N x (0, l), and in particular must be homotopy-equivalent to the compact manifold N. (For a locally compact Hausdorff space X, a neighborhood of infinity may be defined to be the complement of a compact set. An end may be defined to be a connected component of PX \ X, where fix is the Stonetech or maximal compactification of X (the space of maximal ideals of the

7. Another application: the Wall finiteness obstruction

53

algebra of bounded continuous real-valued functions on X). Equivalently, an end is an equivalence class of components of neighborhoods of infinity. In the present situation, the ends must be in one-to-one correspondence with the components of 8W.) So the homotopy type of N is determined by that of the corresponding neighborhoods of the associated end of X. In particular, we now derive a number of necessary conditions for our being able to complete X to a compact manifold with boundary. X must have finitely many ends, and for each end E of X, if Xi is a sequence of connected open neighborhoods of E with Xi \ E, the fundamental groups of the Xi must stabilize to some finitely presented group nl(E) (in the sense of the Mittag-Leffler condition, that l@ rr(Xi) = rl(E), and for each i, the images in ri(Xi) of the ri(Xj), j 2 i, eventually stabilize). Let R = Zrl(E). Then we obtain an inverse system (H.(Xi; R)) of homology groups with local coefficients in R which must also stabilize to what will correspond to H.(N; R). Thus for a suitable open connected neighborhood U of E, ~1 (U) = ~1 (E) and H. (U; R) looks like the homology of a compact manifold of dimension n - 1. If for instance xl(E) is finite (this is not so essential but it already covers an interesting case), R is Noetherian and the homology must be finitely generated by Theorem 1.7.13. By the same Theorem, the cellular chain complex of U with coefficients in R is homotopy-equivalent to a complex of projective modules of finite type. Thus the obstruction g(C.(U; R)) is defined and must vanish in l&(R). Siebenmann’s Theorem says that once this is satisfied, one can put a boundary on the end E provided at least that n # 3, 4, or 5. See [Weinberger, $5 1.5 and 1.61 for a further explanation. The homeomorphism class of the boundary to be added is not always uniquely determined; but the nonuniqueness is also related to K-theory: it is classified by the Whitehead torsion invariant to be studied in 52.4 below. 1.7.15. Remarks. Since this is not a book on topology, we will not prove any purely topological results here. However, in the interests of completeness, let us say a few more words (without proofs) about Wall’s original results (in the topological setting) and about one other important area of application, the spherical space form problem. Wall’s work on the finiteness problem was motivated in part by earlier work of Swan [Swanl] on the question of when a finitely dominated space X, with finite fundamental group 7~ = xi(X) and universal cover X homotopy-equivalent to a sphere, can be homotopy-equivalent to a finite CW-complex. Swan already realized that at least in this particular situation, an obstruction in Kc(&) plays a fundamental role, and he showed how to kill off this obstruction in order to solve a particular geometric problem in which he was interested. This geometric problem was a modified version of what is now known as the spherical space form problem: to classify compact manifolds M”, known as spherical space forms, having a sphere as universal cover, A? 2 S”. Certain obvious examples, such as real projective spaces and lens spaces, arise from free orthogonal actions of finite groups, and the groups that can act in this way are completely known

54

1. KO of Rings

(see [Wolf]). However, a rather subtle question remains: can there be any examples of spherical space forms not homotopy-equivalent to examples of this type, for instance with fundamental groups (such as the non-abelian group of order pq, where p and q are distinct odd primes with pJ(q - 1)) that cannot have a free orthogonal action on a sphere? The answer to this latter question turns out to be “yes,” and the question of what finite groups can act freely on spheres is now totally understood (see [Madsen]). The relevance of the finiteness obstruction comes from the following method of attack. We begin by looking at n-dimensional CW-complexes X with the desired finite fundamental group 7r, having the property that the universal cover X is homotopy-equivalent to a sphere S”. This means of course that X must have vanishing homology in degrees 0 < j < n, and infinite cyclic homology in degree n, but in fact, by the basic theorems of homotopy theory (the Hurewicz and Whitehead Theorems, to be discussed in $5.1 below), this homology condition is not only necessary but also sufficient. The study of the homology of X shows then that r must be a “group with periodic homology” [CartanEilenberg, pp. 357-3581, for which there is an elegant classification theorem [CartanEilenberg, Ch. XII, 5111. It turns out that a necessary and sufficient condition for the existence of such a space X is that each Sylow subgroup of r be either a cyclic group or a generalized quaternion group. However, one is still faced with another problem: given the X whose universal cover is homotopy-equivalent to a sphere, is X homotopy-equivalent to a (smooth) compact manifold M? If the answer is “yes,” then the universal cover of M will be a compact manifold homotopy-equivalent to a sphere. By known results on the Poincare Conjecture, the universal cover is then actually homeomorphic to a sphere, except perhaps when n = 3. (See the remarks following Theorem 2.4.4 below.) A detailed sketch of how this problem is attacked may be found in Madsen’s survey [Madsen]. However, a crucial first step already understood by Swan comes from the well-known fact that any smooth compact manifold is homotopy-equivalent to a finite CW-complex. Thus if M is to exist in the homotopy type of X, the CW-complex X must have vanishing finiteness obstruction in &,(Zr). This group is known to be finite when 7r is finite, but is usually quite hard to compute. For cyclic groups of prime order, we began the calculation of this group (in terms of number-theoretic invariants) in Example 1.5.10, and will complete the calculation in Example 3.3.5(b) below. 1.7.16. Exercise (Nontriviality of the finiteness obstruction for bounded complexes of free modules), Let R be a ring with unit, P a finitely generated projective R-module which is not stably free. (a) Show that there is an R-module homomorphism cp : F + F, where F is a free R-module of countable infinite rank, that is, split surjective with kernel 2 P. (This is attributed to Eilenberg, though the idea may be older. Compare Exercise 1.1.8.)

7. Another application: the Wall finiteness obstruction

55

(b) Deduce that the complex . . . +o+o+F~F+O+O-_,... is homotopy-equivalent to the complex . . . -+o~o+P+o+o+o+..., but not to a complex of finite length consisting of finitely generated free modules. 1.7.17. Exercise. Show that the condition that R be Noetherian in Theorem 1.7.13 is necessary, by exhibiting a non-Noetherian ring R and a chain complex of finite type of free R-modules that does not have finitely generated homology. Hint: find a non-Noetherian commutative ring R (not an integral domain) containing an element x whose annihilator in R is not finitely generated. 1.7.18. Exercise (Behavior of the finiteness obstruction under products). It is of interest to know how homotopy finiteness of X and of X x 2 are related, when 2 is itself a finite CW-complex, for instance, a sphere or a projective space. The algebraic analogue of this is to take the tensor product of two complexes to obtain a double complex. Note that ?ri(X x 2) g 7ri(X) x ri(Z), so that the relevant ring for the geometrical problem is i&(X x 2) 2 @Q(X) x Ti(Z)] = ;zTi(X) @z Zn1(2>.

(4

Show that if (Ci’, d’) and (I?,‘, d’) are complexes of projective Rmodules and S-modules, respectively, then cj = @I c;-l, @zC,2, k=-cc

( dj = d1 @ id + (-l)Pid @ d2

(b)

(cl (4

on 12’; I& Ci

defines a complex of projective R @IZ S-modules, called the total complex of the double complex Ci 8~ Cz. Show that if (C,‘, d’) is homotopy-equivalent to (Ci, &) and if (C,“, d2) is homotopy-equivalent to (C.“, d2), then the total complex of Ci 8~ C,2 is homotopy-equivalent to the total complex of C! @Z Cz. (You can either carry the homotopies around, or else use a mapping cone argument and Theorem 1.7.7 to reduce to the case where one of the complexes is contractible.) Deduce that if X(C’) and X(C”) are well defined, so is X(C), and that if either X(C’) or X(C2) vanishes, so does X(C). Suppose S = Z (this is the algebraic analogue of taking 2 to be simply connected in the geometrical problem). Show that when X(C’) is well defined and C2 is of finite type, then

J?(C) = X(C’)x(C”L

56

1. Ko of Rings

where X(C”) E Kc(Z) = Z. The topological version of this exercise shows that if X is dominated by a finite CW-complex and Z is a I-inite complex with X(Z) = 0, then X x 2 is homotopy-equivalent to a finite complex, and that if 2 is simply connected with X(Z) # 0, then the Wall obstruction of X x 2 is x(2)x (the Wall obstruction of X). In particular, taking a product with S’ kills finiteness obstructions, and taking a product with S2 multiplies them by 2. 1.7.19. Exercise (Algebraic finite domination) [Ranicki]. Recall that a space X is called finitely dominated if up to homotopy it is a retract of a finite CW-complex; in other words, if there is a finite CW-complex Y a n d t h e r e a r e m a p s f :X-+Y,g:Y-+Xwithgof zidx. Nowif X is literally a retract of a finite CW-complex, in other words, if we can arrange to have g o f = idx, then obviously the singular chain complex of X is a direct summand in the singular chain complex of Y, which in turn is homotopy-equivalent to the cellular chain complex of Y, which is of finite type. Thus in this case it is clear that the singular chain complex of X satisfies the hypothesis of Theorem 1.7.12. However, it is perhaps not immediately apparent that the same holds true if we only have go f N idx , for then the singular chain complex C, of X is only a direct summand of the singular chain complex D. of Y “up to homotopy.” The following trick for dealing with the general case is due to Ranicki.

0)

Suppose C. is a chain complex of projective R-modules, bounded below (say non-zero only in non-negative degrees) which is a “direct summand up to homotopy” of a complex of finite type D, of free R-modules. In other words, we assume we are given chain maps f : C. -+ D. and g : D, + C., as well as a chain homotopy h satisfying idc - g o f = d o h + h 0 d. Note that f 0 g - (f o g)" = do f hg + f hg o d, so that f hg gives a chain homotopy between fog and (f o g)2. Show that the endomorphism p of @&, Di given by the matrix

.. .

(2)

.. .

..

is an idempotent, so that its image is a finitely generated projective module over R. Let Ci = @i=, Dj and define a map d’ : C,! -+ C,!_, by the (i-l) xi matrix

7. Another application: the Wall finiteness obstruction

57

if i is even,

if i is odd. Show that (d’)2 = 0, so that (CL, d’) is a chain complex. (3) Define maps cp : Ci * Ci and $ I Ci + Ci by

and by

= hig(zo)

+ hi-‘g (xl) + . . . + hg(zi-1) + g(xi) E Ci.

Show that cp and 1c, are chain maps and that they give a chain homotopy equivalence between C. and CL. (Hint: 1c, o cp = g o f, which we already know is chain homotopic to the identity. The homotopy between cp o $ and the identity is given by a simple “shift” map.) (4) Suppose D. is of “dimension n,” in other words, that Di = 0 for i > n. Thus Ci z @y=, Dj for all i 2 7~. Show by “truncating” CL that its finiteness obstruction (and thus the finiteness obstruction of C.) is well defined, and equal to the class in I&(R) of the image of p from (1).

58

1. Ko of Rings

1.7.20. Exercise [Swanl, $61. The work of Swan discussed in Remarks 1.7.15 above leads to some interesting examples of finitely generated projective modules over group rings. Suppose G is a finite group of order n and let R = ZG, the integral group ring. Define the norm element of R bY iv = I&G g. Observe that for any g E G, gN = Ng = N, so N is central in R and N2 = nN. Let r E Z be prime to n, and let P, be the ideal of R generated by r and N. (It doesn’t matter whether one takes the ideal to be one-sided or two-sided, since N and r are both central.) Obviously PI is just R itself. (1) Show that P,. is the universal R-module defined by two generators u and v and the relations gv = v all g E G, Nu = TV. (Here u corresponds to r and v corresponds to N.) (2) Show that P,. ” P,, provided T = r’ mod n. (Use (1) and define the isomorphism by v H v’, u H ZL’ + hv’, where T - T’ = hn.) (3) Show that R CB P+ Z P,. CB P,I. Note the suggestive analogy with Lemma 1.4.11! (Again use (1). If u” and v” are the generators of P,.,! , send (0, v”) H (v, 0), (1, 0) H (u, au’ + W), and (0, ‘1~“) I-+ (T’21, C(W’ - T’V’)), where a, b, c E Z are suitably chosen.) (4) Choose T and T’ in (3) to be multiplicative inverses of each other mod n, and deduce that P, CB P,I Z R2, hence that P, and P,I are projective modules whose images in l&(R) are the negatives of each other. In particular, we find that if n = 8 and T = 3, then since 3’ = 1 mod 8, P3 defines an element of &-,(R) which must be either trivial or of order 2. It is known to be of order 2 and to be a generator of &(R) when G = Qs is the quaternion group. The projective modules P, naturally arise in the study of the finiteness obstructions coming up in the spherical space form problem (as explained above in 1.7.15).

2 IT1 of Rings

1. Defining K1 Mosccourses in linear algebra begin with a discussion of vector spaces and dimension, and then go on to a study of automorphisms of vector spaces, i.e., linear transformations and their invariants (determinants, canonical forms, and so on). The usual development of K-theory for rings follows the same pattern. One begins by studying projective modules and their stable classification via Ke, and then goes on to the study of the stable classification of automorphisms of free and projective modules, in other words, to invariants of (invertible) matrices, which are given by the functor Kr. We will begin with the classical approach to KI via matrices, and in the next chapter will describe a more category-theoretic approach via the study of the category of finitely generated projective modules. 2.1.1. Definition. Let R be a ring (with unit). Recall the definitions of M(R) and GL(R) from 1.2.2. We call an n x n matrix elementary if it has l’s on the diagonal and at most one non-zero off-diagonal entry. More precisely, if a E R and i # j, 1 < i, j 6 n, we define the elementary matrix eij(a) to be the (n x n) matrix with l’s on the diagonal, with an a in the (i, j)-slot, and with O’s elsewhere. The subgroup of GL(n, R) generated by such matrices is denoted E(n, R). Via the usual embedding of GL(n, R) in GL(n + 1, R) (see 1.2.2), E(n, R) embeds in E(n + 1, R). The infinite union of the E(n, R) is denoted E(R), and is usually called (by slight abuse of language) the group of elementary matrices. The following lemma, which summarizes some easy matrix identities, is only needed in part at the moment, but is included here for future reference. 2.1.2. Lemma. The elementary matrices over a ring R satisfy the relations

2. K1 of Rings

60

eij(u)eij(b) = eij(u +

(a)

b);

ei.j(a)ekl(b) = ekl(b)eij(a), e+j(a)ejk(b)eij(a)- ‘e eij(a)eki(b)eij(a)-‘eki(b)-’

j # Ic and i # 1;

@I

eik(ab), i,j, k distinct; = ekj(--ba),

i,j, k

distinct.

(c)

Cd)

Furthermore, any upper-triangular or lower-triangular matrix with l’s on the diagonal belongs to E(R). Proof. The relations axe easily checked by matrix multiplication. Suppose A = (uij) E GL(n, R) is upper-triangular with l’s on the diagonal. Then A’ = (~$1 = &2(-a1&23(-a23). 3. en--l,n(-an--l,n)

still is upper-triangular with l’s on the diagonal and has O’s on the superdiagonal j - i = 1. Let A” = (a;) = A’e 13(-ui3)e24(-ui4) ...e,-z,n(-uk_2,,).

This now is upper-triangular with l’s on the diagonal and has O’s on the super-diagonals j - i = 1, 2. Continuing by induction, we construct a sequence A, A’, A”, . . . , A(“- l) of matrices in GL(n, R), each differing from the previous one by an element of E(n, R), each upper-triangular with l’s on the diagonal, and with the additional property that uii) vanishes for 0 < j -i 5 k. Thus A(“-‘) = l,, the n x n identity matrix, so A E E(n, R). The lower-triangular case is similar. 0 2.1.3. Corollary. For any matrix A E GL(n, R), the 2n x 2n matrix (:

Aq

lies in E(2n, R).

Proof. Apply the identity

(; Aq =(: :)(-1-l

F) (: :) (! -b)

from the proof of Lemma 1.5.4. By Lemma 2.1.2, the first three factors on the right lie in E(2n, R). And

(Y Tl’)=(i ;‘)(: i)(:, ;‘)r hence the last factor on the right is also in E(2n, R), by Lemma 2.1.2 again. Cl

1. Defining Kl 61

2.1.4. Proposition (Whitehead’s Lemma). For any ring R, the commutator subgroups of GL(R) and of E(R) coincide with E(R). In particular. E(R) is normal in GL(R) and the quotient GL(R)/E(R) is the

maximal abelian quotient GL(R),b of GL(R)I Proof. Since E(R) C GL(R), [E(R), E(R)] c [GL(R), GL(R)]. fir-

thermore, relation (c) of Lemma 2.1.2 shows that

provided i, j, and k are distinct. Thus each generator of E(R) is a commutator of two other generators and [E(R), E(R)] = E(R). We need only show that [GL(R), GL(R)] C E(R). Let A, B E GL(n, R). We embed GL(n, R) in GL(2n, R) and compute that

ABA-IB-l 0

0

AB

1

0

B-F&) (a’ 1) (B;l ;)-

By Corollary 2.1.3, all the factors on the right lie in E(2n, R), so

ABA-lB--l E E(R). 0 2.1.5. Definition. If R is a ring (with unit), we define Kl(R) to be GL(R),b = GL(R)/E(R). Note that R * ICI(R) defines a functor from rings to abelian groups, for if cp : R + 5’ is a (unit-preserving) ring homomorphism, cp induces a map from GL(R) to GL(S) and hence from GL(R),b to GL(&t,. If A, B E GL(R), the product of the corresponding classes [A], [B] E ICI(R) may be represented in two convenient ways. On the one hand, [A] - [B] = [AB]. On the other hand, one may form the “block sum” , and since

Corollary 2.1.3 shows that

[A @ B] = [AB @ l] = [AB]. One may also interpret K1 (R) as the group of canonical forms for invertible matrices over R under elementary row or column operations (in the usual sense of linear algebra). For if A E M(n, R), eij(a)A is the matrix obtained from A by adding a times the j-th row to the i-th row (an elementary row operation), and Aeij(a) is the matrix obtained from A by adding a times the i-th column to the j-th column (an elementary column operation). Vanishing of K1 (R), for instance, would mean that every matrix in GL(R) can be row-reduced or column-reduced to the identity matrix.

62

2. K1 of Rings

2.1.6. Exercise: behavior of K1 under Cartesian products. Let R = RI x Rz, a Cartesian product of rings. By using the obvious decomposition GL(R) = GL(R1) x GL(Rz), show that Kl(R) 2 Kl(R1) x Kl(R2). Generalize to arbitrary finite products. (Compare Exercise 1.2.8.) 2.1.7. Exercise: a ring with vanishing K1. Let k be a field and let V be an infinite-dimensional vector space over k. Let R = Endk(V). Show that K1 (R) = 1. Hint: V is isomorphic to an infinite direct sum of copies of itself. Thus if A E GL(R), one can form

and regard it also as an element of GL( R). Show that A 63 (w. A) is conjugate to (00 . A), hence that A represents the identity in K1 (R). (Compare Example 1.2.6.) 2.1.8. Exercise: Morita invariance of K1. In analogy with Theorem 1.2.4, show that Kl(M,(R)) E Kl(R), for any ring R and any positive integer n. 2.1.9. Exercise: KI of a direct limit. Show by an argument somewhat similar to the proof of Theorem 1.2.5 that if (Ra)aE~, (0,~ : R, --+ Ro)~ 0. (Split into local rings and use Exercise 2.1.6 and Corollary 2.2.6.) 2.2.8. Exercise. Compute Ki(lc[t]/(P)), for any field k and for any integer m > 0. 2.2.9. Exercise (Another approach to a determinant over the quaternions). Let W be the usual ring of quaternions a + bi + cj + dk, where a, b, c, d E Iw and ij = k, i2 = j2 = k2 = -1. Recall that one defines a+bi+cj+dk=a-bi-cj-dk.

4 b)

Show that if one defines N(z) = zZ, then N gives a surjective homomorphism W x -+ rW:. In particular, the commutator subgroup of lHlx must lie in the kernel of N. Show that the kernel of N is exactly the commutator subgroup of IHI’. (Hint: show that iej’i-’ = e-j’, and similarly with i, j, k cyclically permuted. Deduce that e2je1, e2jB2, and e2je3 are all

70

2.

K1 of Rings

commutators. Show that these generate an open neighborhood of 1 in N-‘(l) z S3. But S3 is connected.) Thus RI&, g RG. c) Since W is a vector space over R of dimension 4, W may be embedded in Md(R) by the left regular representation, and G,&(W) +. GL4,(R). Composing with the determinant gives a homomorphism detn : GL,(W) --) Rx. Relate this to the Dieudonne determinant and to N, a n d s h oNw: K1(W) t h -% a tlit:. 2.2.10. Exercise (Some rings of interest in operator theory). Here is an exercise dealing with some rings (actually, algebras over C) of great importance in operator theory and functional analysis. While they are not themselves local rings, we will study a “determinant” somewhat similar to that which we have constructed above in Theorem 2.2.5, and we will make a connection with local rings in the next exercise. Let ‘H be an infinite-dimensional separable Hilbert space with an orthonormal basis ei , e2, . . . . A bounded operator on ‘H is called compact if it sends the unit ball to a pre-compact set, or equivalently, if it is a limit (in norm) of operators of finite rank. It is a well-known fact that the spectrum of any compact operator consists of 0 and of a sequence of eigenvalues tending to 0. (This is to be interpreted to mean “counting multiplicities,” in the sense that no non-zero eigenvalue has infinite multiplicity. Zero itself may or may not be an eigenvalue.) A compact normal operator is diago nalizable. We denote by K(E) the Banach space of all compact operators with the operator norm:

This is a closed twosided ideal in the algebra L3(‘H) of all bounded operators. Now if S is a positive bounded operator on 7-t, its trace is defined by

TrS = e(Sei, ei) E [0,

co].

i=l

The trace is independent of the choice of orthonormal basis, for if the sum converges and el, , ei, . . . is another orthonormal basis, then

g(Sei, ei) = i=l

2 (2 (Sk, e;) eg, ei) 2 (Sei, eg)(eg, ei) i=l

j=l

=

i,j=l

=

5

i,j=l

(Se;, ei)(ei, e[i)

2. K1 of division rings and local rings

=

c(c: a3

03

j=l

i=l

71

Se;, ei) ei, e$ 1

An immediate consequence is that if U is a unitary operator, TrS = Tr(U*SU). (Compute the trace of U*SU using the original basis and the trace of S using the basis {Uej}.) If S = T*T is a positive operator and the trace Tr(S) is finite, then (Sei, ei) + 0, i.e., IITeil) + 0, so that {Tei} is norm-convergent to 0. Thus T is compact and S is compact. If 1 5 p < 00, the Schatten pclass of ‘FI is the Banach space LP(1-I) of operators T for which IITIIP =

Pm7”))+ < 007

where ITI = (T*T)i. The Schatten classes consist of compact operators since this condition implies ITIP is compact, hence ITI and T are compact. It turns out that II lip is a norm and that CP(‘FI) is complete in this norm. Furthermore, L?‘(R) C CJ”(7-L) for p 5 p’, since (for T compact) T E LP(3.t) if and only if the sequence of eigenvalues of ITJ lies in 1P. When p = 2, llTl[~ = (Tr (T’T )); = fJT*Tei, ei) = g(Tei, Tei), i=l

so lITI = (T, T) HS,

i=l

where the inner product ( , )HS is defined by

(T, S)HS =

C(Tei, Sei). i=l

Thus in this case L2(7-Q is a Hilbert space, called the Hilbert space of Hilbert-Schmidt operators. In general note that clearly llXTllp = lXlllTllp and lITlIp > 0. If lITlIp = 0, then the positive quadratic form defined by ITIP vanishes on all the ei, hence everywhere, so ITIP = 0, IT( = 0, and T = 0. The triangle inequality can be verified by showing first that lITlIP =

sup

F of finite rank IIFIIqIl

I TvF)l7

P

Q

where if p = 1 we interpret IIFIIq to mean the operator norm of F. (Since TF has finite rank, its trace is well defined in the usual sense.) One can also check easily that f?‘(N) is a two-sided ideal in L3(‘FI) (though not closed in the operator norm). The space ,Cl(‘FI) is called the space of trace-class operators. If T E L1 (‘H ), the sum C(Tei, ei) converges absolutely, and defines a linear functional TrT independent of the choice of orthonormal basis (just as before). Hence, once again Tr(U*TU) = Tr(T) for U unitary.

72

2. K1 of Rings

Now let k(3-1) = C.l~+K(‘H), and similarly let LP(X) = C.lx+LP(‘FI). Each of these rings has a unique maximal two-sided ideal, of codimension one. (For instance, X(1-t) is a two-sided ideal in X(7-I) of codimension one, so it is maximal even as either a left ideal or right ideal.) (1) Complete the proof that 1) )I 1 is a norm and that L’(X) is complete, by showing that for A E 8(7-l) and T E C’(a),

Hint: Use the polar decomposition T = U(T( to split AB a~ (AU)TI+)(JT\j) anduse the Cauchy-Schwarz inequality for ( , )HS. Then if T, S E ,!Z’(l-t), write T + S = U)T -I- SI (polar decompo sition) and estimate Tr(lT + Sl) as Tr(U*(T + S)) = Tr(U*T) + Tr(V*S) via the above estimate. (2) Show that if T or S is of trace class, then Tr(TS) = Tr(ST). Hint: if T is of trace class and S is unitary, this follows from invariance of the trace under conjugation by S. Now get the result for all S (with T still of trace class) by taking linear combinations. Show that t:(X) and f?(X) have split surjections onto @ inducing (3) surjections on Kr. (4) (The operator determinant) Let R = k’( N), the trace-class operators with identity adjoined. Let RF = ker(RX + C’), and call this the group of determinant-class operators. Construct a homomorphism det : Rr -+ Cx with the property that (*)

det(eT) = eTrcT)

for T E L’( X).

(Here the exponential of an operator is constructed via the usual exponential power series.) Hint: First show that every determinant-class operator D is an exponential of a trace-class operator. One can do this by noting that every element of the spectrum of D, except perhaps for 1, is an eigenvalue of finite multiplicity, and that 1 is the only accumulation point of the spectrum. Hence, if VI is the span of the generalized eigenspaces for D corresponding to the eigenvalues X with IX - 11 2 1, one obtains a (not necessarily orthogonal) direct sum decomposition of 3-1 into two invariant subspaces VI and V, for D, where VI is finite-dimensional and the spectral radius of (D - 1)/v, is < 1. Then one can take a logarithm of D(v, using the usual power series

and choose any logarithm for the invertible operator D(v, of finite rank (using, say, the Jordan canonical form). Next, observe that if T and S are both of trace class and eT = es = D, then if T has eigenvalues Aj and S has eigenvalues pk, the

2. K1 of division rings and local rings

73

set {exj} must coincide with the set {el-lk}, and the multiplicities must match up. On the other hand, Xj + 0 and jJk + 0. One can see from this that again one can find a (not necessarily orthogonal) direct sum decomposition of l-l into two invariant subspaces VI and Vz for both T and S, where VI is finite-dimensional and eTlvl = eslvl, and where TI v, = Slv,. In particular,

n(T) --T’(S) = ‘l l(Tlv,) -ll(Sl,) E %iZ,

so eTrtT) = em(‘) .

This shows that (*) gives a well-defined definition of det. Finally, show that the determinant is multiplicative, i.e., that if T and 5’ are of trace class, then det(eTeS) = det(eT) det(es). On can do this using the Campbell-Baker-Hausdorff formula etTesS -_

and the fact ((2) above) that Tr vanishes on commutators. (5) Extend the definition of det to a homomorphism defined on ker[GL(R) + GL(C)]. (Hint: if T E GL(n, R) and T H 1 E GL(n, C), then T may be viewed as a determinant-class operator on 7f @c C”.) 2.2.11. Exercise (A local ring in operator theory). In this exercise, we pursue the use of K-theory in operator theory in the context of local rings. Let 3-t be a complex Hilbert space as in the last exercise and let A be some algebra of bounded operators on ‘H, not necessarily with unit. Thus A could be B(X) or Ll(7-t). Let R be the ring of formal operator-valued power series a0 . 1 + aA + z2A2 + -. 1, where Aj E A for j 2 1 and the constant term a0 . 1 is a scalar multiple of the identity operator. Show that if a0 # 0, then a0 . 1 + zAl + z2A2 +. . . has an inverse in R. Deduce that R is a local ring, with radical the power series without constant term. (2) If A is a Banach algebra, show that the same holds for R’ if we define R’ similarly using only those power series with a positive radius of convergence in z, in other words, with germs at z = 0 of analytic operator-valued functions in place of formal power series. Let A E 23(X). Then 1 - zA has an inverse in R, which is essen(3) tially (except for the change of variable z I--+ z-‘) what is called in operator theory the resolvent of A. Show that the power series for (1 - zA)-’ converges for Iz( < IIAl(-l. Let A = L’( R). Show that the determinant of the last exercise (4) defines a homomorphism from (R’) ’ to the group of units in the commutative local ring of germs of analytic functions around 0.

(1)

74

2. K1 of Rings

Show also the following useful fact: if A is a trace-class operator, f~(z) = det(1 - zA)-’ extends to a function of z analytic in the whole complex plane except perhaps for countably many isolated singularities, and that if zc is a zero or singularity of f~, then ~0’ E Spec.4. (Actually, more is true; f~ is entire analytic, and f~(zc) = 0 if and only if ze-’ E Spec 31 for more details.)

3. K1 of PIDs and Dedekind domains As we did in Chapter 1 in studying Kc, we shall proceed from the study of Ki of division rings and local rings to the study of Kr of the most elementary examples of non-local commutative rings. Of particular interest are the sorts of rings that occur in algebraic geometry and number theory. Here we shall discuss PIDs and Dedekind domains; polynomial rings will be dealt with in the next chapter. The easiest examples to treat are Euclidean rings. These include Z, the Gaussian integers Z[i], Z[w], the rings of integers in a few other special number fields, and the polynomial ring Ic[t] in one variable over a field Ic. To fix notation, we remind the reader of the basic definition. 2.3.1. Definition. A (commutative) integral domain R is called a Euclidean ring or Euclidean domain if there is a norm function ) 1 : R + N with the following properties: (i) If a E R, [a( = 0 if and only if a = 0. (ii) If a, b E R, (ubl = lullbl. (iii) (Euclidean algorithm) If a, b E R, b # 0, then there exist q, T E R, called the quotient and remainder, respectively, such that a = qb + r and 0 5 (T] < (bJ. In the examples Z, Z[i], Z(w], and k[t], the norm function is given by the usual absolute value, by (a + &i( = o2 + b2, by (a + b*I = u2 - ub + b2, and by (f(t)]=2desf (with the convention that deg0 = --co), respectively. 2.3.2. Theorem. If R is a Euclidean ring, then SKI(R) vanishes and Kl(R) g RX. In fact, for each TX, SL(n, R) = E(n, R). Proof. Let A = (uij) E GL(n, R). We try to proceed roughly as in the proof of Proposition 2.2.2, but the problem is of course that there is no guarantee that there will be an invertible entry in a given row or column of A. However, the norm function on R gives us a mechanism for doing an induction. To illustrate, start with the first column of A. Not all elements of this column can be zero, so there is some air # 0 and with Iair ) minimal subject to this condition. If Jail] = 1, then ai1 must be a unit. (By the Euclidean algorithm, 1 = quil + r with 0 _< ]rl < 1, hence with Jr] = 0, so r = 0 by (i) of (2.3.1).) If lair] > 1, then ai1 is not a unit, and so

3. K1 of PIDs and Dedekind domains

75

generate s a proper ideal (ail). On the other hand, since A is invertible, the ideal generated by the elements of the first column must be all of R, and so there is some j # i with ajl $ (ail). Applying the division algorithm giveg ajl = &&I + r, where jr/ < [ail). Since uji $ (ail), T # 0 and thus jr/ > 0.

So by subtracting q x (i-th row of A) from the j-th row, we can row-reduce

A to decrease the minimal norm of a non-zero element in the first column.

Once we’ve shown this, then iterating the reduction procedure enables us to reduce to the case where there’s a unit in the first column. So then we can proceed as in the case of R a field and row-reduce A to the form a11

*

where ali is a unit and A’ is of size (n - 1) x (n - 1) and ( 0 A’ > ’ invertible. Then we repeat the whole process with A’, etc. The rest of the proof is identical to that of Proposition 2.2.2. •I 2.3.3.

Corollary.

K1(Z) = (1, -l}, Ki(Z[i]) 2 (1, i, - 1 , -i},

Kl(Z[v]) g {6-th roots of l}, and Ki(lc[t]) % Icx.

Proof. In the examples of 2.3.1, it’s easy to see which elements have norm 1. q Theorem 2.3.2 naturally raises the question of whether the same statement is true or not for more general PIDs or Dedekind domains. Unfortunately, the answer is “no”; there are PIDs with non-zero SlYi, though they are not so easy to find. (For examples, see [Ischebeck] and [Grayson].) Thus it seems the idea of the proof of Theorem 2.3.2 cannot be pushed any further. However, there is one general result about Ki of Dedekind domains that arises as a special case of Bass’s general theory of “stable range.” One may view the vanishing of SK1 for a commutative ring R as the statement that Kl(R) is generated by the image in GL(R) of GL(1, R). When this doesn’t hold, the next best thing would be for Kl(R) to be generated by the image in GL(R) of GL(2, R). Instead of trying to explain the general theory (for which one can consult [Bass]), which gives for a ring R an estimate on the smallest value of n for which K1 (R) is generated by the image in GL(R) of GL(n, R), we will give a simplified proof of the one case we need. We begin with a lemma which will also be used in Section 5 of this chapter. Because of Corollary 2.1.3, Lemma 1.5.4 is just a special case of the following. 2.3.4. Lemma. Let R be a ring (with unit) and I a two-sided in R. Then for any n, the natural map E(n, R) -+ E(n, R/I) is surjective. Proof. By definition, E(n, R/I) is generated by elementary matrices eij(b), where b is the image in R/I of a E R. Such a matrix clearly lifts to the elementary matrix eij(u) E E(n, R). q

Kl(R) is genera by the image in GL(R) of GL(2, R) (in fact, by the images in GL(R) of GL(1, R) and of SL(2, R)).

76

2. Kl of Rings

Proof. Let A E GL(n, R) and suppose n > 3. We will show that A 1 * can b% row-reduced to a matrix of the form , where A’ is of size 0 ( A’ > ( n - l ) x ( n - l )and invertible. Subtracting aii x (the first column of A) 1 0 from the i-th column then reduces A to the form o B with B E ( > GL(n - 1, R), so [A] E ICI(R) lies in the image of GL(n - 1, R). Induction on n then gives the result of the theorem. (We already know the image of GL(2, R) is generated by GL(l, R) and by SL(2, R).) Now consider the first column of A. Since A is invertible, the ideal generated by its entries is all of R. We will show we can do elementary row operations on A to put at least one zero in the first column. One this is done, the ideal generated by the remaining entries in the column is all of R, so adding multiples of the other rows to the row with the zero, we can change the zero to a 1. Then if necessary, we may premultiply by eii(l)eii(-l)eii(l) to put the 1 in the (1, 1)-slot. Subtracting multiples of the first row from the other rows then reduces A to the desired form. Let I be the ideal generated by usi,. . . , a,~. If I = 0, then asi = 0 and we’re already done. If I = R, then subtracting a linear combination of rows 3 through n from the first row puts a zero in the (1, 1)-slot, and we’re again done. So we may assume I is a proper non-zero ideal. By Theorem 1.4.7, we may factor I uniquely into a product of maximal ideals. By the Chinese Remainder Theorem, this gives a corresponding factorization of R/I into a product of local rings of the form R/P”, where P is a maximal ideal. By Proposition 2.2.4, SK1(R/Pk) = 0, so by Exercise 2.1.6, SKl(R/I) = 0. In fact, by the method of proof, we know that SL(m, R/I) = E(m, R/I) for any m. We will use this fact for m = 2. For each element a E R, let b be its image in R/I. Since Rull + . . . + Run1 = R, dividing by I gives that (R/I)cill + (R/I)& = R/I. In other words, we can find xi and 22 in R such that &iLii + ?z&i = i, or det(221

2i) =I.

So we have a matrix in SL(2, R/I) = E(2, R/I). By Lemma 2.3.4, it lifts to an elementary matrix ( fi2

if) in SL(2, R), and blxl + b2x2 = 1

(here we may have to change the original x1 and x2 within their I-cosets). But on the other hand, xiuii +x2uzi - 1 E I, so there exist x3,. . . ,x, E R with c&i ziuii = 1. For i 2: 3, we have xi = xi(blxl + b2x2). So we get the equation xi011

+x2a21+ (x3blxlax

+x3b2xmd+. . .+ (xnb1x1unl +x,bzx2u,l) = I

or xi(oii + xsblusi + . . . + x,blu,l) +

x2(u21+

x3bs-m +. . . + xnbsu,l) = 1 .

This says exactly that by adding (x3bl) x (the 3rd row) + . . f + (xnbl) x (the nth row) to the first row, and by adding (x3b2) x (the 3rd row) +. . -+

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(z,bs) x (the yth row) to the second row, we can change A so that the ideal generated by the new all and ~21 is all of R. Then subtracting a linear combination of the first and second rows from the last row, we can achieve the desired zero. Cl The above theorem suggests studying, for commutative rings R and especially Dedekind domains, the subgroup of SKE(R) generated by the image of SL(2, R). The convenient way to do this is in terms of so-called Mennicke symbols. 2.3.6. Theorem. Let R be a commutative ring. For a, b E R with Ra + Rb = R, choose c, d E R with ad - bc = 1. z i E SL(2, R) isindependent of ( ) the choice of c and d, hence can be denoted [a b] without possibility of confusion. Such an element of SKI(R) is called a Mennicke symbol, and if R is a Dedekind domain, all elements of SKI(R) are of this form. [a b] = 1 ifae RX, bE R. For a, b E R relatively prime, the Mennicke symbols satisfy the relations [a b] = [b u] and [a b] = [a + bX b] for any X E R. If Rala2 + Rb = R, then [al b] . [a2 b] = [0102 b]. Then the class in SK1 (R) of

Proof. (1) The assertion that the class of

is independent of the

choice of c and d follows immediately from the calculation that if

,

E SL(2, R), then

(; 1) (; ;)-‘=(; ;) (!;I ;p)=(,,‘,, ;>. The Mennicke symbols clearly exhaust the image of SL(2, R) in Kl(R), so by Theorem 2.3.5, they exhaust SKI(R) if R is a Dedekind domain. (2) is clear from the fact that if a E RX and A =

(

z

1

>

E SL(2, R),

then we can subtract ca-’ x (1st row) from the second row to change A to the form : o!r

(

. Then multiplying by the elementary matrix )

a-l 0 makes the matrix strictly upper-triangular, hence elementary. ( 0 a> For (3), note first that (: :) (:I :)=(I: $7 so [u b] = [-b u]. When we verify (4), it will follow that [a b] = [-b a] = [b a][-1 u] = [b u]

(by (2)).

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Furthermore,

so [u b] = [a+bX b]. To check (4)) assume Ralaz + Rb = R. Then if have determinant 1,

1 0 0 now keeps the first row the same and 0 0 1 0 -1 0 puts a 1 in the (3, 3)-slot, and further elementary operations reduce the matrix to the form ala2 b 0 * o . ; 01 ) ( Premultiplying by

( 1

S o [ai b][az b] = [a102 b]. 0 2.3.7. Corollary. If R is a Dedekind domain and R/P is a finite field for each non-zero prime ideal of R, then SKI(R) is a torsion group. Proof. Consider a Mennicke symbol [o b]. If b = 0, then a E RX so [a b] = 1 by (2) of the theorem. Similarly, [a b] = 1 if b E RX. If neither is the case, (b) is a non-zero proper ideal of R and so is a product of non-zero primes ideals Pj by Theorem 1.4.7. Since each R/Pj is finite, it follows that R/(b) is finite (cf. the beginning of the proof of Theorem 1.4.19). Since the image of a in R/(b) is a unit and (R/(b)) ’ is a finite group, there is some k with uk E 1 mod (b), and then by (4) of the theorem, [cz b]” = [u” b] = [l + bX b] by (3) and then (2) of the theorem.

(for some A)

= [l b] = 1

0

This is about as much as one can say about general Dedekind domains. However, for the examples of greatest interest in number theory, namely the rings R of algebraic integers in number fields (finite extensions of Q), it turns out that one can explicitly compute RX and also show that SKI(R)

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a real vector space of dimension ri + r2 - 1. Now a bound on

implies a bound on the absolute values of the elementary symmetric functions of the ~j (a), which are the coefficients of a manic polynomial equation satisfied by a, and are ordinary integers. So the inverse image under X of any given ball of lRTl+r+2 is finite, which shows that X(RX) is discrete and the kernel of X is finite. The kernel of X therefore consists of a E R for which oq = 1 for some q, in other words, of roots of unity. On the other hand, since A maps into a torsion-free group, all roots of unity in F must lie in the kernel of A, and the kernel coincides with the group of roots of unity in F, the torsion subgroup of RX. If F = Z, then obviously RX is just {fl}, and coincides with the kernel of A. If F is an imaginary quadratic field, then rs = 1, ~1 = 0, and V = 0, so again RX = ker A. Furthermore, for general F, since the image of X is a discrete subgroup of a real vector space of dimension ri + 7-s - 1, X(RX ) is free abelian of rank < ri + r:! - 1, and RX is finitely generated. It remains only to show that the rank of X(RX) is precisely ~1 + ~2 - 1. obvious elements of RX other than the roots of unity, even if ~1 + r2 - 1 V , is large. Since X(W) X(RX) h a s r a n k e q u a l t o it sh e e dq iumi ve to showing that V/X(RX) is compact, or to showing that there is some compact subset K of V whose translates under X(RX ) cover V. To show this, we first recall that by the proof of Theorem 1.4.18, o f tadditive group h e of R as a lattice rry;” 0.j (discrete cocompact subgroup) in lR r1 x F'. In particular, the volume (in the sense of n-dimensional Lebesgue measure) of (RF x P)/a(R) is some finite positive constant, say Ci.TJ = (WI,. . . , u,,+,~) E V, l e t

Note also that since Cuj = 0, the product of the coordinates evj is 1 Hence e” . a(R) (where . denotes coordinatewise multiplication) is again a lattice in lP x cCr2 of covolume Ci. So if Q is a closed cube or ball of volume > Ci centered at the origin in llV x Cc'*, its image in the quotient by e”.a(R) must have smaller volume, hence there had to be two points 21 and x2 in Q with the same image. In other words, 21-22 E e”.a(R), so that 2Q (the cube or ball with dimensions twice as big) contains a point of e” .0(R). Let K’ be the compact image of 2Q under the map lW’~ x P -+ lRrl+rz defined by taking the logarithm of the absolute value of each coordinate. Then we have shown that for all points 2, E V, v + A(R \ (0)) meets K ’. This is almost, but not quite, what we want, since we are interested in X(RX), not X(R \ (0)) (wh’ic his a semigroup but not a group). However, if C’s denotes the maximum L1-norm of a point in K’, in other words, the

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maximum value of the sum of the coordinates, then any a E (R \ (0)) with e” . a(a) E 2Q must satisfy

However, as observed in the proof of Theorem 1.4.19, there are only finitely many integral ideals in R of norm 5 ecz (for any Cz), and so up to units there are only finitely many possibilities for a, say al,. . . , arc. Thus we have shown that for any v E V, there is a unit u E RX such that v+ A(aj) +X(U) meets K’ for some j 5 k. Thus there is a compact set K independent of v such that v + X(U) meets K for some u E RX, and this proves the theorem. (Take K = U3=1(K’ - X(uj)).) q It has been shown in [BassMilnorSerre] and in [Milnor, $161 that in fact SKI(R) vanishes when R is the ring of algebraic integers in a number field, so that Theorem 2.3.8 gives the complete calculation of Kl(R) in this case. However, this is not an easy theorem and there doesn’t seem to be an elementary proof. With less effort, one can prove somewhat less, for instance, that SK1 (R) is finite. There are quite a number of proofs available, though all seem to require some additional tools. One method is to first show that SL(2, R) is finitely generated, for instance, by constructing an explicit fundamental domain for SL(2, R) as a discrete subgroup of a product G of ri copies of SL(2, R) and of rp copies of SL(2, C). It then follows from Theorem 2.3.6 and Corollary 2.3.7 that SKI(R) is finite. An alternative argument in [Kazhdan] uses representation theory. One can show that for each n, SL(n, R) is a discrete subgroup of a product G(n) of ri copies of SL(n, IFi) ando rs f copies of SL(n, Cc), and that the quotient G(n)/SL(n, R) has finite invariant measure. On the other hand, Kazhdan shows that for n L 3, the locally compact group G(n) has property T, i.e., its trivial one-dimensional representation is an isolated point in the space of all irreducible unitary representations of the group. Kazhdan also observes that property T inherits to discrete subgroups of cofinite volume and to quotients thereof. Therefore the abelianization SL(n, R)at, has property T. However, for a locally compact abelian group A, the irreducible unitary representations are just the continuous homomorphisms into T, the circle group, so property T means that A = Hom(A, T) is discrete. For A discrete, A is compact, so the only way it can also be discrete is if it is finite. So SL(n, R)ab is finite for n > 3. In particular, SKI(R), which we have seen is a quotient of SL(3, R)at,, is finite. 2.3.9. Exercise (Finite generation of E(n) and SL(n)). (1) Show using Lemma 2.1.2(a) that if a ring R is finitely generated as a Z-module, then E(n, R) is finitely generated as a group. Deduce from Theorem 2.3.2 and Corollary 2.3.3 that SL(n, Z), SL(n, Z[i]), and SL(n, Z[=-%$]) are finitely generated groups for all n. (This is not so easy to show directly.) (2) Show using Lemma 2.1.2(c) that for any ring R, E(n, R) is its own commutator subgroup (i.e., is a perfect group) for n 2 3.

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Also use Lemma 2.1.2(c) to strengthen the result of (1): if a ring R is finitely generated as a Z-algebra, then E(n, R) is finitely generated as a group for n 2 3. (3) Show that SL(2, Z) = E(2, Z) is not, its own commutator subgroup, by exhibiting a homomorphism onto an abelian group. Hint: what is SL(2, Z/(2))? 2.3.10. Exercise (Stabilization of GL(n)/E(n) domains).

for Dedekind

(1) Let R be any ring. Show using the proof of Proposition 2.1.4 that [GL(2, R), GL(2, R)] C E(4, R) (when GL(2) is embedded in GL(4) as usual). (2) Again let R be any ring. Show that the image of GL(2, R) in GL(n, R) normalizes E(n, R) if n 2 3. Hint: first note that the image of GL(2, R) normalizes the subgroup El generated by the eij(u) with i 5 2 and j 2 3, the subgroup El generated by the eij(a) with j 5 2 and i 2 3, and the subgroup Es generated by the eij(a) with i, j 2 3. Then use Lemma 2.1.2(c) to show El, E2, and E3 generate all of E(n, R). (3) Now let R be a Dedekind domain. By the proof of Theorem 2.3.5, if n 2 3, GL(n, R) is generated by E(n, R) and by the image of GL(2, R). Deduce from this fact and from (1) and (2) above that for any n 2 3, E(n, R) is normal in GL(n, R), and that for any n 2 4, GL(n, R)/E(n, R) is the abelianization of GL(n, R). (In fact there are cases where E(2, R) is not normal in GL(2, R). With somewhat more work, one can show that GL(n, R)/E(n, R) is already abelian for n = 3.) (4) Deduce from (3) and from part (2) of Exercise 2.3.9 the following theorem about finite generation of SL(n, R): if R is a Dedekind domain which is finitely generated as a Z-algebra, and if SL(2, R)at, is finitely generated, then SL(n, R) is finitely generated as a group for all n 2 4. (As remarked in (3), this can be strengthened to n 2 3.) 2.3.11. Exercise (Non-triviality of Mennicke symbols). The following famous example from [BassMilnorSerre] shows there are Dedekind domains with non-trivial Mennicke symbols. Let R = lk[x, y]/(cc’ +y2 - l), the ring of polynomial functions on the circle. This is a Noetherian integral domain with field of fractions F = lR(z, y)/(s2 + y2 - 1). (1) Show that R is a Dedekind domain. (This part of the exercise also appeared in Exercise 1.4.23. There are several possible arguments, such as checking the original definition or showing that R is integrally closed in F and applying Theorem 1.4.17.) (2) Observe that

and that for any n 2 2, the

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83

associated function S1 -t SL(n, R), defined via the formula

represents a non-trivial element of nl(SL(n, R)) g rl(SO(n)) (see Example 1.6.13 for the calculation of this fundamental group). (3) Argue on the other hand that if g(x, y) E E(n, R), then the matrix-valued function (x, y) I-+ g(x, y) E SL(n, R) must represent 0 in rl(SL(n, R)). Hint: it’s enough to check this for elementary matrices, for which there’s an obvious homotopy to a trivial loop. (4) Deduce that there’s a homomorphism SKI(R) --+ K?‘(S’) = Z/(2) sending [x y] to the non-zero element of Z/(2). (5) Show that in fact [x y] is an element of order 2 in SKI(R) by using Theorem 2.3.6 to show [x y12 = 1.

4. Whitehead groups and Whitehead torsion For applications of K1 to topology, just as in the case of the Wall ob struction, the rings of interest are integral group rings ZG, where G is a group which in the applications is the fundamental group of some topological space. Note that Kl(ZG) always contains certain “obvious” elements, namely the images of the units fg, g E G. We therefore focus attention on the “non-obvious” part of K1 (ZG). 2.4.1. Definition. If G is a group, its Whitehead group Wh(G) is the quotient of K1 (ZG) by the image of {fg : g E G} s (ZG)X. Thus if G is the trivial group, Wh(G) = Kl(Z)/{fl} is trivial by Corollary 2.3.3. The rings ZG are in general quite complicated from the ringtheoretic point of view; for instance, in what would appear to be the simplest non-trivial case, if G is the cyclic group of two elements with generator t, the map a+bt H (a+b, a-b) embeds ZG into the Cartesian product ZXZ as what we called in Definition 1.5.1 the double D(Z, (2)) of Z along the ideal (2). The units fl, ft of ZG correspond in D(Z, (2)) to f(1, 1) and to f(1, -l), which are all the units of Z x Z, so Wh(G) = SK1 (D(Z, (2))). One can show that this vanishes (see Theorem 2.4.3 below), but to do this from scratch is a bit involved, and this only handles the case of the simplest non-trivial group! Thus the computation of Whitehead groups is usually not easy. Nevertheless, the Whitehead groups of finite groups are now thoroughly understood, and we refer the reader to [Oliver] for a complete treatment. Here we content ourselves with a few elementary results. Since it may not be apparent from Definition 2.4.1 that Whitehead groups are ever non-zero, we begin with an example.

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2.4.2. Example. Let G be a cyclic group of order 5, with generator t. We shall exhibit an element of infinite order in Wh(G). Let a = 1 - t -t-l and note that (1 - t - t-l) . (1 - t2 - P) = I- t - t-l - t2 + t3 + t - t3 + t-1 + t2 = 1, so that a E (ZG) x. Under the homomorphism Q : ZG + @ defined by sending t ++ e2rri/5, {fg : g E G} maps into the roots of unity and in particular into the complex numbers T of absolute value 1. So b I--+ [o(b)/ defines a homomorphism from Wh(G) to lRT. Since la(a) 1 = 11 - e2xi/5 - e-2xi/5 I = 11 - 2 cos T I x 0.4, we deduce that (I: gives an element of infinite order in Wh(G). The example may be generalized. Suppose G is any group and we are given a homomorphism a : G -+ U(n), the unitary n x n matrices over Cc. This group homomorphism clearly extends to a ring homomorphism cy : ZG + A&(C), and thus induces a homomorphism a* :

K~(ZG) -+ Kl(iK(c:)) 2

Kl(c) g

cx.

(Here we have used Morita invariance, Exercise 2.1.8.) But a(fG) C u(n), which maps to T in Ki(C) under the determinant. Hence the absolute value of the determinant gives a homomorphism a, : Wh(G) + R$ which can be used to detect elements of infinite order in the Whitehead group. Detecting elements of finite order in Wh(G) is trickier and requires more sophisticated methods. Nevertheless, the technique of Example 2.4.2 in fact detects all of Wh(G) for many groups of practical interest, for instance for cyclic groups, though we aren’t prepared to prove this at the moment. To give an idea of what can be done by brute force, we show that the Whitehead group of a cyclic group of order two is trivial. (More powerful methods of computation use the exact sequences of the next section and Chapter 4.) 2.4.3. Theorem. The Whitehead group of a cyclic group of order two is trivial. Proof. We have seen above that this is equivalent to proving that SK1 (D(Z, (2))) vanishes. Suppose (A, B) E SL(n, D(Z, (2))). This means A, B E SL(n, Z) and A - B = 0 mod 2. By Theorem 2.3.2, A E E(n, Z). Thus clearly (A, A) E E(n, D(Z, (2))). Multiplying (A, B) by (A, A)-‘, l,, the n x n identity ma trix. So suppose A = 1, and B = 1, mod 2. If we could row-reduce B = (bij) to the identity matrix by elementary operations involving adding even multiples of one row to another row, then it would be clear that (1, B) E E(n, D(& (2))).

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85

So we try to apply the division algorithm as in the proof of Theorem 2.3.2. Let (&I,.. . , b,l) = (bl). Then bi is even and bii is odd. We show that we can reduce B by elementary operations of the allowable sort so fl * that bii = 51, bi = 0, i.e., B = Then we repeat the same B’> ’ ( 0 procedure with B’, and so on. Eventually we come down to the case where B is upper-triangular with fl’s on the diagonal and even entries above. More allowable elementary operations now reduce B to a diagonal matrix with fl’s on the diagonal, and since det B = 1, the number of -1’s is even. To finish the argument, we only have to see what to do with the case n=2, B = (_d -4) (since after renumbering of the rows and columns B is a direct sum of blocks of this type and of some identity matrix). In

fact the matrix ( -b_

-4 ) is not contained in the subgroup of SL(2, Z)

generated by (i T) andby (i f

);

however,

((; f), (_d _4))=(1’-lM1,-Q-1 is elementary as a matrix over D(Z, (2)) by Corollary 2.1.3. So this completes the argument except for the step about reducing bii to fl and bi to 0. For this we note that if (bill = 1, we can subtract even multiples of the first row of B from the other rows and thereby reduce jbil to 0. If (bl( = 0, then since @ii) + (bi) = Z, we must have Ibill = 1. If \bli( > 1 and lbrl > 0, there are two cases, depending on which of these is larger. If Ibill < Ibi(, then by the division algorithm we can write br = qbll + T with 0 < (r( < /bill (T- can’t be 0 since b ii and bi are relatively prime). If q is even, then we may reduce the size of lbil by adding even multiples of the first row to the other rows. If q is odd, then r is odd and we write instead bi = (q f l)bii + (r F bri). With the correct choice of the sign, we have 0 < Ir F bill < Jbril, but q f 1 is even so we can argue as before. In the other case, IbiiJ > Jbil. Again we apply the division algorithm and obtain bii = qbl + T with 0 < (T( < (bi( and r odd. If q is even, this means we can subtract even multiples of other rows from the first row to reduce the absolute value of bii. If q is odd, we use the same trick as before and write bii = (q f 1)bi + (r F bi) with the sign chosen so that 0 < Ir F bil < lbil. Again we can subtract even multiples of other rows from the first row to reduce the absolute value of bii. After repeating the algorithm finitely many times, we eventually come down to the case where Ibill = 1. 0 The reader will presumably agree after seeing this proof that computing Whitehead groups from scratch is not very practical. But at least we know now that Wh(G) is trivial for some finite groups and infinite for others.

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2. ICI of Rings

In the rest of this section, we will give a brief exposition of the concept of Whitehead torsion, which provides the motivation for introducing the Whitehead groups. Whitehead torsion gives an algebraic obstruction for homotopy equivalences between certain topological spaces to be “simple,” or of the “obvious” sort. Since for present purposes a homeomorphism is to be viewed as an “obvious” sort of homotopy equivalence, Whitehead torsion can be used to distinguish homotopy-equivalent spaces which are not homeomorphic. The most famous application of Whitehead torsion is the “s-cobordism theorem,” which is the main tool in classifying manifolds in dimension > 5. So that the reader can appreciate the importance of the Whitehead groups for topological problems, we will give the statement here. However, we shall not discuss the proof as it will take us too far afield. For details, see [MilnorHCT] for the simply connected case and [RourkeSanderson, Ch. 6] and [Kervairel] for the general case. Theorem (“s -cobordism theorem ”- Barden, Mazur, 2.4.4. Stallings). Let M” be a connected compact n-manifold of dimension 2 5 with fundamental group X, and consider the family 3 of all “h-cobordisms” built on M. These are connected compact manifolds Wn+l with exactly two boundary components, one of which is M” and the other of which is some other manifold M’“ , such that W has deformation retractions onto both M and M’. There is a map r : 3 + Wh(r), called the “Whitehead torsion,” and r induces a natural one-to-one correspondence from 3’/ N to Wh(n), where N is the equivalence relation induced by homeomorphisms W + W’ which are the identity on M. If W is the “t rivial” h-cobordism W = M x [0, 11, then r(W) = 1. 2.4.5. Corollary. If Mn is a connected compact n-manifold of dimension 2 5 with fundamental group x, and if Wh(r) = 1 (for instance, if M is simply connected or if r is of order 2), then every h-cobordism built on M is homeomorphic (rel M) to a product M x (0, 11. In particular, the other boundary component M’ is homeomorphic to M. Remarks. We have been deliberately vague about what category of manifolds we are dealing with here. In fact, the theorem is valid in all three of the major categories of manifolds: topological manifolds and continuous maps, PL manifolds and PL maps, and smooth manifolds and C” maps. In the last of these, “homeomorphism” in the theorem is to be interpreted as “diffeomorphism.” One of the main applications of the Corollary, as noticed by Smale, is the proof of the PoincarC conjecture: that in dimension n 2 6 (this can be reduced to 5 with a little more work), any manifold C” homotopyequivalent to 5’” is (topologically) homeomorphic to S”. Furthermore, the set of diffeomorphism classes of smooth homotopy spheres Cn is in one-to-one correspondence with the group Diffe(S”-l) of isotopy classes of diffeomorphims of S”-l. To prove this, cut out two small disks from C”, viewed as the “polar caps” of the homotopy sphere. What remains is a manifold W” with the homotopy type of a cylinder and with two

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87

boundary components each homeomorphic to S*-‘. Since n - 1 2 5 and S”-l is simply connected, the hypotheses of the Corollary are satisfied and there is a homeomorphism (or diffeomorphism, if C is a smooth manifold) from W to S”-l x [0, l] which is the identity on the boundary component corresponding to the south polar cap. Hence we can glue the south polar cap back in and deduce that Cn % B” Uf Bn, a union of two balls glued by a homeomorphism (if we’re in the topological category) or diffeomorphism (if we’re in the smooth category) f from S”-1 to itself. In addition, it’s clear that any such f defines a homotopy sphere B” Uf B”. The equivalence class of this homotopy sphere only depends on the isotopy class of f, since an isotopy of f’s gives an h-cobordism of the corresponding homotopy spheres and we can apply the Corollary again. Conversely, if there is an orientationpreserving diffeomorphism from B” Uf Bn to the standard sphere, it is not hard to see that there must be an isotopy from f to the identity. This explains why the smooth homotopy spheres are parameterized by Diffe(S”-l). In the topological category, since B” is the cone on Sn-‘, any self-homeomorphism f of S”-l extends to a self-homeomorphism F of Bn by the simple formula

F(T2) = rf(z),

r E [O, 11, 2 E F-l.

(This is the “Alexander trick.“) This yields a homeomorphism from B” Uf to S”, proving the Poincare conjecture. The most elementary context in which to discuss “simplicity” of homotopy equivalences is that of a finite relative CW-complex (X, A). In other words, we assume A is a (Hausdorff) topological space and that X is obtained from A by attaching finitely many cells, so that k-cells are always attached before (k + 1)-cells and the inclusion A - X is a homotopy equivalence. We assume as well that A and X are both path-connected and locally simply connected, with the same fundamental group 7r (computed with respect to some basepoint zs in A). Let X and A be the universal covers of X and A, which carry free actions of 7r by covering transformations, and let R = Z&r be the group ring of *IT. In this situation, the relative homology groups H.(X) A; Zr) = H.(_%, A; Z) must vanish. However, these may be computed from the cellular chain complex C.(X) A; i&r) = C.(_%, A; Z), which is the direct sum of one free rank-one R-module in degree k for each k-cell added in obtaining X from A. The hypothesis that A - X is a homotopy equivalence means (by the Whitehead and Hurewicz theorems) exactly that this chain complex of finite type is acyclic. The Whitehead torsion of the homotopy equivalence will be an invariant of the chain complex C.(X) A; Z&r) defined using one extra piece of structure-a choice of basis elements for the free modules Ck (X, A; R). Since the k-chain module contains one free rank-one R-module for each geometric k-cell, there is a choice of a basis which is canonical up to an element of {*g : g E n} for each cell. Namely, we choose a basis element for the free cyclic submodule corresponding to each cell in X \ A, and it only depends on a choice of orientation for this cell (hence the f sign) and

B”

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on a choice of a lift of this cell to a cell in X \ A (hence the element of the covering group). If there are only cells in two consecutive dimensions, k - 1 and k, then once we have fixed our basis elements, the differential dk : Ck(X, A; R) --+ &-1(X, A; R) must be given by an invertible n x n matrix over R, where n is the number of k-cells or (k - 1)-cells. (The number of cells must be the same in both dimensions since H.(X, A; Q) must vanish, hence dimCk_r(X, A; Q) = dimCk(X, A; Q).) 2.4.6. Definition. The Whitehead torsion Q-(X, A) of the homotopy equivalence A -+ X is the image in Wh(n) of the matrix of dk in GL(n, R) if k is even, or the inverse thereof if k is odd. Note that while the matrix of dk is not well defined as it depends on the choice of basis, the torsion is well defined since we have divided out by all possible ambiguities. Now consider the general case where C.(X, A; R) is allowed to be any acyclic chain complex of finite type of free R-modules, starting in degree 0, with bases chosen for each chain module. By the argument in the proof of Theorem 1.7.12, one may increase the ranks of the chain modules (adding “cancelling pairs” of cells in consecutive dimensions) so that Bk = zk =&f kerdk is free for each k. Then dk defines an isomorphism Ck/Bk + Bk_i. We choose bases for the non-zero Bk’s, taking the basis for Bo = Co to be the basis we already have for Co, and idempotents pk : ck --+ Bk. Then pk @3 d,+ : Ck --+ Bk ‘$9 Bk_1 is given by an invertible matrix with entries in R, and we let [dk] be its class in Wh(n). (We can suppress the pk because if pi is another projection from Ck onto Bk, then pk -p$ vanishes on Bk and hence factors through dk. But the matrix of (pk + s o dk) @ dk differs from that of pk @ dk by an elementary matrix, so their classes in Ri (R) are the same.) The Whitehead torsion T(X, A) of the homotopy equivalence A L) X is then defined to be the alternating product (since we’re writing Whitehead groups multiplicatively) n, [dk](-l)k. This is independent of the choice of bases for the Bk’s, since if we change the choice of basis for Bk by an invertible matrix P, this multiplies the matrix for dk+l by P and the matrix for dk by P-l, so that we get cancellation in the alternating product. Notice also that this agrees with our previous definition when C, = 0 for p # k, k - 1, since Bk = 0 and Bk-r = C&i, so that we can use the same basis for Bk-r as for C&r. Note finally that the fact that we had to stabilize to make all the Bk’s free, by adding on “cancelling pairs” of cells in consecutive dimensions, does not matter, since this kind of geometric stabilization corresponds to passage to the limit from GL(n, R) to GL(R) in the definition of K1. There is a geometric definition that corresponds to the algebraic condition of vanishing torsion. 2.4.7. Definition. The homotopy equivalence A of X is called elementary, or given by an elementary collapse, written X Le A, if X is obtained from A by attaching two cancelling cells in adjacent dimensions; in other words, if for some k, X = (AUfBk-l)U,B”. Here f : Skp2 -+ A is the attaching map for the (k - 1)-cell and we suppose g : Sk-l + (Auf Bkwl)

4. Whitehead groups and Whitehead torsion

89

maps one hemisphere identically onto the (k - 1)-cell and the other hemisphere of Sk-l into A. This is illustrated in the following picture.

k-l

2.4.8. Figure: An elementary collapse Note that if f collapses Ske2 to a point a, this just means that X = AV,Bk and one can obviously collapse Bk to the attaching point a. In the general case, f extends to a map f : Bkml + A and X has a deformation retraction down to A collapsing the k-cell down to f(B”-l), as one can see in Figure 2.4.8. More generally, we say X collapses to A or A expands to X and write X \ A or A /’ X if and say the homotopy equivalence A - X is simple if it is in the equivalence relation generated by \, i.e., if X /” X1 \ X2 /” . . . \ A (with all the collapses and expansions fixing A pointwise). 2.4.9. Theorem (Geometric characterization of Whitehead torsion). In the above context of a finite CW-pair (X, A) with A and X Hausdorff, path-connected, and locally simply connected, and where the inclusion A L) X is a homotopy equivalence, the inclusion is simple if and only if T(X, A) = 1 in Wh(r). In particular, if Wh(7r) = 1, for instance if X and A are simply connected or r is of order 2, then every such homotopy equivalence A - X is simple. firthermore, for fixed A and a fixed element (Y E Wh(r), there exists a finite CW-pair (X, A) such that the inclusion A L) X is a homotopy equivalence with 7(X, A) = a.

Proof (Sketch). If X Le A, then 7(X, A) = 1 since the boundary map in the cellular chain complex just corresponds to the 1 x 1 matrix (l), as one can see from Figure 2.4.8. Next observe that if

90

2. K1 of Rings

and all the inclusions are of finite CW-pairs and are homotopy equivalences, then r(X, A) = T(X, X,_l).+~~(X1, A). This follows from the fact that

C.(X, A; R) = C.(X, X,-I; R) @...@C.(Xl, A; R) and the matrix defining [&I for (X, A) differs from the direct sum of those defining the [dk] for the successive pairs (Xj, X,-l) by an elementary matrix. It follows that the torsion vanishes if X \ A. The same principle also shows the torsion vanishes if A of X is simple, for if for instance X1 \ X 2 A and Xr \ Xs 2 A, then 7(X1, A) = 7(X1, X)7(X, A) = 7(X1, X2)7(X2, A), so T(X, A) = 7(X2, A). The general csse follows from the same argument by iteration. The existence part of the theorem is a direct construction. Given Q E Wh(r), realize it by a matrix B E GL(n, R). Then let X1=AvP2v./S2, n

times

and construct X from Xr by attaching n 3-cells so that in the universal cover the cellular boundary map is given by

B : C3(X, A; R) 2 Rn + R” 2 C2(X, A; R). This is possible since ~(zr, A) is a free R-module on n generators. Then (X, A) obviously has the right torsion. For the last part of the theorem, one needs to note first that if X’ differs from X by a homotopy of the attaching maps for the cells rel A, then X can be converted to X’ by a sequence of expansions and collapses (rel A). For this it’s enough to consider the case of X = A Uf,, B” and X’ = A Ufl B”, where f : Sk-’ x [0, l] + A is a homotopy of attaching maps. Merely define W = A Uf (B” x [0, l]), which is defined by attaching Bk x [0, l] to A along Sk-l x [0, 11. Then (W, A) is a finite CW-pair: one can first attach two k-cells to A via fe and fr , then glue in a (k + 1)-cell Bk+l G Bk x [0, 11 via f on Sk-’ x [0, l] and via the identity maps to the two k-cells along Bk x (0, 1). But W Le X and W Le X’ since one can “cancel” the (k + 1)-cell with either of the two k-cells. The hardest part of the theorem is to show that if r(X, A) = 1, then A L) X is simple. For this the idea is to proceed in two steps: first to modify X (rel A) by means of elementary expansions and collapses (which as we have seen do not affect the torsion) so that all the cells added to A to form X are in two consecutive dimensions k and k - 1, then to show that each elementary matrix operation applied to

dk : Ck(X, A; R) g R” + R” g C&1(X, A; R)

4. Whitehead groups and Whitehead torsion

91

has a geometric analogue. Here we only deal with the last part; see [RourkeSanderson] or [Cohen] for the full argument. Suppose Xr is obtained from A by attaching n (Ic - 1)-cells, and X is obtained from Xi by attaching n k-cells via an elementary matrix eij(o). Using the observation about homotopies of attaching maps, one can change X by expansions and collapses so that for m # j, the m-th &cell is glued onto the m-th (Ic - 1)-cell as in Figure 2.4.8, and the pair of cells collapses down to A. The j-th k-cell is glued onto both the j-th (Ic - 1)-cell and the i-th (k - 1)-cell. But now since the i-th (Ic - 1)-cell can be collapsed down to A (along with the i-th k-cell glued onto it), the attaching map for the j-th k-cell can be homotoped through A to “unhook” this cell from the i-th (k - 1)-cell. So after further expansions and collapses, we can assume each k-cell is glued onto exactly one (Ic - 1)-cell as in Figure 2.4.8, and the cells can be collapsed in pairs down to A. Cl The concept of Whitehead torsion can be carried over from inclusions A L) X to general homotopy equivalences f from one finite (connected) CW-complex Xi to another, Xs. To do this, if f is cellular, form the mapping cylinder X = Cf = Xi x [0, l] Uf X2 (here we use f to attach Xr x { 1) to X2). Since we assumed f is cellular, this is a finite CW-complex, and since f was assumed a homotopy equivalence, it has deformation retractions down to the subcomplexes A = X1 x (0) and X2. We define r(j) = T(X, A). Note that if f is actually an inclusion of a finite CW-subcomplex, then the pair (X, A) is an expansion of the pair (X2, Xl) and so r(f) agrees with our existing definition of r(Xs, Xl). Furthermore, if two homotopy equivalences fe and fi : Xi --+ Xs are homotopic to one another, then Cf, is obtained from CfO by a homotopy of attaching maps, and hence by the proof of Theorem 2.4.9, their torsions are the same. This makes it possible to define r(f) for a homotopy equivalence f which isn’t cellular. We homotope f to a cellular map fc (this is possible by the “cellular approximation theorem”) and define 7(f) = r(fc). The result is well defined since if we homotope f to a different cellular map fi , then fc N fi and so r(fr) = r(fe). It also turns out that if f is a homeomorphism, then r(f) = 1, but this is a hard theorem [Chapman] unless f is cellular, in which case it’s a triviality. (If f is a cellular homeomorphism, then Cf is cellularly isomorphic to Xr x [0, 11, which clearly collapses to Xr .) For further discussions of the various guises and applications of Whitehead torsion, see [MilnorWT] and [Weinberger, Ch. 11. 2.4.10. Exercise. Extend the proof of Theorem 2.4.3 to show that the Whitehead group of any elementary abelian 2-group (product of finitely many cyclic groups of order 2) is trivial. 2.4.11. Exercise (Behavior of Whitehead torsion under products). This exercise is in some sense the Kr-parallel of Exercise 1.7.18. (a) Suppose (C,‘, d’) and (Ci, d’) are complexes of finite type of based free R-modules and S-modules, respectively, with Ci acyclic (so that r(C,‘) is defined). Show that the total complex of the double

92

2.

Kl of Rings

complex Ci @, 17: of free R C&J S-modules, cj= 6 C;_l,EJ3aC;, k=-co

i d.j = d1 @id + (-l)pid @ d2 on Cp’ @z Ci is also based and acyclic. (b) Suppose that in the situation of (a), 5’ = Z. Show that T(C) = @)x(C2), where x(C”) E Ko(Z) = Z. Cc) Suppose A L) X is a homotopy equivalence satisfying the hypotheses of Theorem 2.4.9, so that its torsion is defined, and let Z be a finite connected and simply connected CW-complex. Show using (b) that A x 2 L) X x 2 is also a homotopy equivalence satisfying the hypotheses of Theorem 2.4.9, and that 7(X x 2, A x 2) = T(X, A)x(Z). Thus if 2 = S3, deduce that A x 2 - X x 2 is always simple. (4 Show also that in the situation of (a), if x(C”) = 0, then 7(C) = 0 regardless of what S is. Deduce that if 2 = S’, then Ax2 - Xx2’ is always simple.

5. Relative K1 and the exact sequence As with Ko, we want to be able to relate K1 of a quotient ring R/I to Kl(R) and to some invariants of the ideal I (and the way it is embedded in R). In this section, we will define the relative group K1 (R, I) and show that the three-term exact sequence of Section 1.5 extends to a six-term exact sequence relating Ko and K1. This will provide us with some more computational tools for computing K-groups. 2.5.1. Definition. Let R be a ring (with unit) and let I be a two-sided ideal in R. We define D(R, I) as in 1.5.1 and define the relative Kl-group of the ring R and the ideal I to be Kl(R, 1) = ker ((PI), : Kl(D(R, 1)) -,

Kl(R)) .

Note that this is the exact parallel of Definition 1.5.3. Since it’s convenient to have another definition closer in spirit to Definition 2.1.5, we now prove a relative version of Whitehead’s Lemma and rework the definition of K1 (R, I) into a more usable form. 2.5.2. Definition. Let R be a ring (with unit) and let I be a two-sided ideal in R. We define GL(R, I) to be the kernel of the map GL(R) -+ GL(R/I) induced by the quotient map R --f R/I. We define E(R, I) to be the smallest normal subgroup of E(R) containing the elementary matrices eij(a), a E I. Note that since each such elementary matrix is congruent to the identity matrix modulo I, E(R, I) C GL(R, I).

5. Relative K1 and the exact sequence

93

2.5.3. Theorem (Relative Whitehead Lemma). Let R be a ring (with unit) and let I be a twosided ideal in R. Then E(R, I) is normal in GL(R, I) and in GL(R),

GL(R, I)/E(R, I) = Kl(R, I), and GL(R, I)/E(R, I) is the center of GL(R)/E(R, I). Furthermore, E(R, I) = [E(R), E(R, 01 = FL(R), EC& 01. Proof. The first assertion follows from the fact that if A E GL(n, R) and B E E(n, R, I), then

(““u”-l y)=(; ;1) (f ;) (a’ ;). Since

is elementary by Corollary 2.1.3 and by its definition (: $1) E(R, I) is normal in E(R), the right-hand side lies in E(R, I). Next suppose (Al, AZ) E GL(D(R, I)) c GL(R x R) and maps to the identity element of Kr (R) under (PI),. This means of course that Al E E(R). But then (Al, Al) E E(D(R, I)), since if Al = &eikj,.(uk),

(Al, Al) = neiljk(akl 4. k

Multiplying (Al, AZ) by (Al, Al)-’ changes it to the form (1, B) with B E GL(R) but without changing its class in Kr. Since (1, B) E GL(D(R, I)), B = 1 mod I and B E GL(R, I). C onversely, every B E GL(R, I) defines a class in GL(D(R, I)). So to show GL(R, I)/E(R, I) 2 Kl(R, I), we need only check that if B E GL(R, I), then (1, B) E E(D(R, I)) if and only if B E E(R, I). For one direction, note that E(R, I) is generated by matrices of the form Seij(a),!Y1 with a E I and S E E(R). But

(1,

Seij(a)S-l)

=

(S, S)eij(O, u)(S-l, S-l)

and all three factors on the right lie in E(D(R, I)). For the other direction, suppose

n

(h B, = fi %jk (uk7 bk) e E(D(R, I)), k=l

eibjh

(uk) = 1 E E(R).

k

Note that for each k,

eikjk(uk7 bk) = eihjk(uk7

uk)eikjk(o, bk

-

uk) = (Sk, Sk)& Tk),

where Sk = %jk(ak) E E(R),

Tk = eikjk(bk - ak),

b k - ak E I.

2. Kr of Rings

94

Then we have k

= (1, (s,T,s,1)(sls,T2s,1s,1) . . . (S,& . . . S,T$y *. . S,lS,l))

)

since Si S2 . . . S, = 1, and we’ve written our element B as a product of generators of E(R, I). Since E(R, I) is normal in GL(R, I) and in GL(R), [E(R), E(R, I)] C [GL(R), E(R, I)] C E(R, I). Equality holds since E(R, 1) is generated by matrices of the form Seij(a)S-’ with a E I and S E E(R), and Seij(e)S-’ = [S, eij(e)]eij(a> = [S, eij(a>][eik(I>, ercj(u>] E [E(R), JW, 41,

k # i,j.

It remains only to show that GL(R, I)/E(R, I) is the center of GL(R)/ E(R, I). Note first that if A E GL(R, I),

and since A - 1 has its entries in I, and

lie in E(R, I), hence this calculation shows that

A 0

So GL(R, I) and GL(R) commute modulo E(R, I). On the other hand, the center of GL(R)/E(R, I) must map (under the homomorphism induced by the quotient map R ++ R/I) to the center of GL(R/I), which is trivial. (A central matrix must be diagonal with equal diagonal entries, but since for a matrix in GL all but finitely many of the diagonal entries are 1, GL(S) has trivial center for any S, in particular for S = R/I.) Hence the center of GL(R)IE(R, 1) is contained in the kernel of the map to GL(R/I), which is GL(R, I)/E(R, I). Cl We’re now ready for the main theorem of this section, which is an extension to the left of the exact sequence of Theorem 1.5.5.

5. Relative K1 and the exact sequence

95

2.5.4. Theorem. Let R be a ring and I C R an ideal. Then there is a natural exact sequence Kl(R, I) + K1(R) 3 Kl(R/I) 2 Ko(R, I) + Ko(R) 3 Ko(R/I), where q* is induced by the quotient map q : R --H R/I and the maps Kj (R, I) + Kj (R) are induced by pz : D(R, I) + R. Proof. For simplicity of notation in the proof, if A is an element of R or a matrix with entries in R, we will often denote q(A), the corresponding matrix over R/I, by A. We begin by proving exactness of Kl(R, I) + Kl(R) 3 Kl(RII).

We have seen that any class in K1 (R, I) is represented by (1, B) E GL(D(R, I)) 2 GL(R x R) with B E GL(R, I), so B = i and q,[B] = 1. Conversely, if B E GL(R) and q,([B]) = 1, then B E E(R/I). Now if b E R/I, it comes from some a E R and eij(&) = q(eij(a)). So each generator of E(R/I) lies in the image of E(R) and hence E(R/I) = q(E(R)) (this argument was used in Lemma 1.5.4). So B lifts to a matrix C E E(R), and q(BC-l) = 1. Then (1, BC1) E GL(D(R, I)) and [B] = [BC-‘1 in Kl(R) comes from [(l, BC-l)] E Ki(R, 1). Next we have to define the boundary map K1 (R/I) -r?, Ko(R, I) and prove exactness at Kl(R/I) and at Ko(R, I). Theorem 1.5.5 will then complete the proof. The definition of the boundary map is based on what in topology is called a “clutching” construction. Given A E GL(n, R/I) (the image of some matrix A E M,(R), not necessarily invertible), we use A to “clutch” together two free modules to get a projective module over D(R, I). In other words, let R” xA R” = { (2, y) E R” x R” : G = k/i}.

(We are thinking of z and y as 1 x n matrices.) Make this into a module over D(R, I) by letting

This makes sense since ii = i2, hence q(r2y) = f2j_j = iI = q(rlz)A.

Note that if A = q(A) with A E GL(n, R), then

(CC, y) ++ (zA, y) E R” Xi R” 2 D(R, I)n

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2. K1 of Rings

sets up an isomorphism from R” XA Rn to a free module of rank n. In particular, since we have seen that E(R/I) = q(E(R)), R” XA R” is free of rank n if A. is elementary. For a general A E GL(n, R/I), we can always.choose B E GL(n, R/I) such that A@B is elementary (for instance, B = (A)-’ works by Lemma 1.5.4 or Corollary 2.1.3), and then (R” xA R”) @ (R” xs R”) E R2” xAeS R2” Z D(R, I)2n, so that R” XA R” is a direct summand in a free module, i.e., a projective

module. Thus it makes sense to define

a[/i] = [R”

xA R”] - [D(R, I)n] E Ko(D(R, I)).

We will show that d is in fact a homomorphism Kr (R/I) + &(R, I). It maps into &(R, I) = ker(pr), since

bd*(+il) = (n)e([Rn

XA

R”]) - (m),([D(R,

I)n]) = [R”] - [R”] = 0.

It is additive on direct sums of matrices since (R” x/i R”) @ (R” xB R”) % R2” xABh R2n,

and it sends classes of elementary matrices to 0 since if A is elementary,

a[A] =

[R” xA R”] - [D(R, I)“] = [D(R, I)n] - [D(R, I)“] = 0.

More generally, it is well defined on classes in Ki since if A = &’ with B E E(R), then

(5, y) H (zB, y) E R” xc. R” sets up an isomorphism from R” XA R” to Rn XC R”. Thus we obtain a well-defined homomorphism Ki (R/I) + &(R, I). Furthermore we have already seen that the composite

is zero. The composite Kl(R/I) : Ko(R, I) + Ko(R)

is zero since (~2)4+41)

= b2L([Rn x~ R"]) - (pz)*([D(R, I)n]) = [R”] - [R”] = 0.

It remains only to check that ker d C q* (K1 (R)) and that

ker{(m)* : Ko(R, I>

4

Ko(R)) G a(Kl(R/I)).

5. Relative K1 and the exact sequence

97

Suppose a([A]) = 0. Th’is means that R” XA R” is stably isomorphic to a free module of rank n, or that for some m, R”

xA

R” @ D(R, I)m = D(R, I)n+m.

After replacing A by A $ l,, we may assume that in fact R” xA R” E D(R, I) “. Choose an isomorphism cp : D(R, I)” = R” Xi R” + R” Xi R”. Then we can define matrices B,C E M,(R) by (ejB, ejC> = ‘p (ej, ej), where ej is the j-th standard basis vector for R”, or in other words by taking the j-th rows of B and C to be the first and second coordinates (respectively) of cp(ej, ej). Then by linearity, cp(u, V) = (uB, UC) for any (u, V) E D(R, I)” = R” xi R”, and since for such u and V, ti = ti, we have BA = 6’. Since ‘p is invertible, it is clear that B and C are invertible with ‘p-‘(z, y) = (zB_‘, yC_‘) for (2, y) E R” XA R”. Thus A = q(B-lC) and so kera C q*(Kl(R)). Finally, suppose one has a class in Ko(R, I) going to 0 in Ko(R). This means we have a class in Ko(D(R, I)) going to 0 under both (pi), and (~2)~. Represent the class by [P] - [D(R, I)n], where P is a projective D(R, Q-module such that (pr ), (P) and (pz)*(P) are stably isomorphic to R”. If necessary, we may add on a free module of rank Ic to P and replace n by n + Ic so that (pi>*(P) and &s)*(P) are both actually isomorphic to R”. Then it is clear that P is of the form R” x A R”, and thus

PI - MR, 4’7 = a([&. This completes the proof. 0 2.5.5. Corollary. (Cf. Exercise 1.5.11.) Let R be a ring, I G R an ideal such that the quotient map q : R + R/I splits (in other words, such that there exists a ring homomorphism s : R/I + R with q o s = idR,I). Then

o -+ K,,(I) --+ Ko(R) -+ Ko(RII) -+ 0 is split exact. Proof. Clearly s* is a splitting for q*, by functoriality of Ko. We need only show that Ko(I) ---f Ko(R) is injective. But this follows from the fact that s* : Kl(R/I) + Kl(R) is a splitting for q* : Kl(R) -+ Kl(R/I), hence d = 0 in the exact sequence of 2.5.4. q

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2. K1 of Rings

2.5.6. Examples. (Cf. Examples 1.5.10.) (a) Suppose R = Z and I = (m), where m > 0. Then ICI(R) g {fl} by Corollary 2.3.3, while Ki (R/I) was computed in Exercise 2.2.7. It is thus possible to compute Ks(1) from the exact sequence. For example, suppose m = 2. Then R/I is the field of two elements and (R/I)X = (1). The exact sequence therefore becomes K1(R, I) + {fl} + (1) 3 K,,(l) --+ Z 5 Z, and J&,(1) = 0. At the same time, we see that Kl(R, I) must surject onto {fl}. Next, suppose m = p is an odd prime. Then R/I is the field lFP of p elements and (R/I) x is cyclic of order p - 1. Hence the exact sequence becomes K1(R, I) + {fl} + “p” 2 Ks(l) --f Z 5 Z, and &(I) g F~/{fl}, which is cyclic of order q. In this case, the map Kl(R, I) -+ {fl} is trivial. As a third example, suppose m = 2” is a power of 2 with T > 1. Then R/I is a local ring with maximal ideal of index 2, and (R/I)X is an abelian group of order 2’-l. Furthermore, fl are distinct elements of this group. For instance, if m = 8, then since any odd square is = 1 (mod S), all elements of (R/I) ’ are of order 2 and (R/I)X is a Klein Cgroup (Z/(2) x Z/(2)). By Corollary 2.2.6, Ki (R/I) E (R/I) ‘. The exact sequence has the form Kl(R, I) --+ {fl} -+ (R/I)X 2 &(I) + Z 5 Z, and &(I) g (R/I)X/{fl}, an abelian group of order ZTW2 which is not necessarily cyclic. Again in this case, the map Kl(R, I) + {fl} is trivial. (b) Suppose G is a cyclic group of prime order p, say with generator t, and R = ZG is its integral group ring, which may be identified with Z[t]/(t* - 1). If 6 = e2rri/*, a primitive pth root of unity, and if S = Z[c], then S is the ring of integers in the cyclotomic field Q(E), hence is a Dedekind domain by Theorem 1.4.18. There is a surjective homomorphism R ++ S defined by sending t +-+ E(n + 1, R, I’) for some non-zero ideal 1’. Show that H > E(R, I’). Then let I be the largest two-sided ideal of R such that H 2 E(R, I). If H $ GL(R, I), let H’ be the image of H in GL(R/I), repeat the same reasoning with H’ 2 GL(R/I), and derive a contradiction. To prove the uniqueness of I, note that if E(R, 1) c H C GL(R, J), then projecting to R/J, we obtain E(R/J, (I+ J)/J) = 1, hence I C J. Thus if E(R, J) C H C GL(R, I) also, J G I and I = J. Deduce from Proposition 2.2.2, from Corollary 2.3.3, and from Exercise 2.5.17 that the Congruence Subgroup Problem has an affirmative answer if R is a field or R = Z. 2.5.22. Exercise (Non-triviality of relative Mennicke symbols). Let R be the Dedekind domain of Exercises 1.4.23 and 2.3.11, i.e., lR[z, y]/ (x” + y2 - 1). It was shown in the second of these Exercises that [X y] represents an element of order 2 in SKI(R). By Exercise 1.4.23, I = (y, 2 - 1) is a prime ideal in R and R/I 2 IK. Deduce from 2.5.10 that SK1 (R, I) # 0, in fact that [Z y]r # 0 in SK1 (R, I). Is this element also of order 2?

3 KO and IS1 of Categories, Negative K-Theory

1. I-Co and K1 of categories, Go and G1 of rings For many of the applications of K-theory, it is useful to have the notion of K-theory for categories and not just for rings. In this more general context, the K-theory of a ring R is just the K-theory of the category Proj R of finitely generated projective modules over R. Another natural example is the topological K-theory of a compact space X, which is the K-theory of the category Vect X of (locally trivial, real, or complex) vector bundles over X. The identification of this with the K-theory of the ring R = C(X) then follows from an equivalence of categories Proj R 2 Vect X. But there are also many examples that don’t come so directly from rings; for instance, if X is a projective algebraic variety, one can consider in a similar way the category Vect X of algebraic vector bundles over X. We will see many more examples shortly. To begin with, we need to place limitations on the sorts of categories we will consider. These are of two sorts. On the one hand, the category needs to have enough structure so that it makes sense to talk about an object as being built up as an extension of smaller objects. There are several ways of ensuring this and we’ve chosen here what seems to be the most standard choice, though not the most general one. In addition, the category has to be “small” enough to avoid set-theoretic difficulties when we try to make isomorphism classes of objects into a group. Of course, it suffices to require that the category be “‘small” in the usual sense of category theory (i.e., for its objects and morphisms to constitute sets), but this seems overly restrictive since the natural examples Proj R and Vect X are not small categories. This should explain the following definition. Call a category A preadditive (this term is not entirely standard) if Horn (A, B) is an abelian group for each A, B E Obj d, and if composition of morphisms is bilinear. Recall first of all that an additive category is a preadditive category A with a distinguished object 0 such that Hom(A, 0) = 0,

1. Ko and K1 of categories, Go and GI of rings

109

Horn (0, A) = 0 for each A E Obj d, equipped with a binary operation @ which is both the categorical product and the categorical coproduct. An abelian category is an additive category in which every morphism has a kernel and cokernel, and in which every monomorphism is a kernel and every epimorphism is a cokernel. Any abelian category has a notion of exact sequences for which the Five-Lemma and Snake Lemma are valid. Good general references on abelian categories are [Mac Lane] and [Freyd], though we will need very little of the theory developed in these books. 3.1.1. Definition. A category with exact sequences is a full additive subcategory P of an abelian category d, with the following properties: (1) P is closed under extensions, i.e., if

is an exact sequence in A and PI, P2 E ObjP, then P E ObjP. (2) P has a small skeleton, i.e., P has a full subcategory PO which is small, i.e., such that Obj PO is a set, and for which the inclusion PO + P is an equivalence. The exact sequences in such a category are defined to be the exact sequences in the ambient category A involving only objects (and morphisms) all chosen from P. 3.1.2. Examples. (1) Any small abelian category, or more generally any abelian category with a small skeleton, is a category with exact sequences. Examples include the category of finite-dimensional vector spaces over a field F, or the category of finite-dimensional complex representations of a topological group G. To get a small skeleton, take {F” : n E N} in the first case, or {Hom(G, GL(n, Cc)) : n E N} in the second case. When G = Z, the category of finite-dimensional complex representations of G may be identified with the category of pairs (V, T), where V is a finite-dimensional complex vector space and T E Aut V is the image of the generator of G. Another similar example is the category of finite-dimensional complex representations of the monoid N, which may be identified with the category. of pairs (V, T), where V is a finite-dimensional complex vector space and T E End V. (2) Let R be a ring. Then Proj R, the category of finitely generated projective R-modules, is a category with exact sequences, with small skeleton the set of direct summands in {R” : n E N}. However, this is usually not an abelian category since the cokernel of a map between projective modules is usually not projective (think of the simple case R = Z, Z -% Z). The category Proj R has the additional property, not true for the category of finitedimensional complex representations of Z, that every short exact sequence splits.

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(3) Let R be a ring and let R-Modf, be the category of finitely generated R-modules. This is an additive subcategory of the abelian category of all R-modules, and has as a small skeleton the set of quotient modules of the {R” : n E N}. If R is not left Noetherian, this is not an abelian category, since the kernel of a map between finitely generated R-modules may fail to be finitely generated. (If I is a left ideal of R that is not finitely generated, then R and R/I are singly generated but the kernel of the quotient map R -+ R/I is not finitely generated, so this morphism doesn’t have a kernel in the category.) Nevertheless, R-Modfa is always a category with exact sequences, since if

is an exact sequence of R-modules with Ml and M2 finitely generated, one can choose a finite set of elements of M whose images in Mz generate M2, and these together with the images of a finite set of generators of Ml will generate M. (4) Let R be a ring and let R-Mod+, be the category of R-modules with a finite-type projective resolution, i.e., R-modules M for which there exists an exact sequence (3.1.3)

0 -+ P, -i . . . +P,+M-+O

with Pj E Obj Proj R. This is a full additive subcategory of R-Modf,, and may or may not coincide with R-Modf,. If it does and R is left Noetherian (so that R-Modf, = R-Mod+, is an abelian category), the ring R is said to be (left) regular. For a ring to be left regular, it is sufficient (but not necessary) that it be left Noetherian and have finite global dimension (which means that there exists an N such that every R-module has a projective resolution of length 5 N). For the fact that R is left Noetherian implies that every finitely generated R-module has a resolution by finitely generated projective modules, and the global dimension condition then guarantees that every such resolution has length 5 N. In particular, any PID is left regular (since any submodule of a free module is free). Any Dedekind domain R is left regular, since R is Noetherian by Theorem 1.4.5, and the proof of Corollary 1.4.6 shows that every submodule of a finitely generated free R-module is projective. The group rings of non-trivial finite groups are not left regular. To see this, note that for a non-trivial finite cyclic group H one has H,(H, Z) # 0 for all odd n, so that the finitely generated ZH-module % cannot have a finite projective resolution. Then if G is any non-trivial finite group, we can choose a non-trivial cyclic subgroup H G G, and it follows from “Shapiro’s Lemma” that H,(G, ZG @L-H Z) 2 Hn(H,

z> # 0

1. Ko and K1 of categories, Go and G1 of rings

111

for all odd n, so that the finitely generated ZG-module ZG @zH Z cannot have a finite projective resolution. We will see in Proposition 3.1.4 below that whether or not R is left regular, R-Modr,,, is a category with exact sequences. (5) Let A be an abelian category in which every simple object is iso morphic to an element of some set S of objects. (A simple object in an abelian category is the natural generalization of a simple module over a ring; it is an object M E Obj A (with M # 0) such that any monomorphism N H M is either 0 or an isomorphism. The definition has a number of immediate consequences. If M is simple, then End A4 is a division ring (Schur’s Lemma), and any non-zero morphism M + M’ is necessarily a monomorphism, since its kernel N H M can’t be an isomorphism, hence must be 0.) Call the simple objects in A objects of length one, and define inductively (for n 2 2) the objects of length n to be those objects M E Obj A for which there is an exact sequence in A

with MI of length n - 1 and with Mz E S. We will see in Prop* sition 3.1.5 below that the full subcategory AR of A consisting of objects of finite length, objects M of length 5 n for some n, is a category with exact sequences. The Jordan-Holder Theorem holds in this context (with the usual proof), i.e., for M of finite length, the length e(M) is well defined, and the simple modules that occur in a “composition series” for M are unique up to isomorphism and permutation. The category of finite-dimensional representations of a (topological) group G is a good example of a category of objects of finite length. (6) Let X be a compact Hausdorff space. Then Vect X is a category with exact sequences, equivalent to Proj R, R = C(X), by Theorem 1.6.3. Here one can work over either B or C. (7) Let X be a projective algebraic variety [Hartshorne, Ch. I, ‘$21 over an algebraically closed field (or more generally a projective scheme--see [Hartshorne, Ch. II]-over a commutative Noetherian ring). Then Vect X, the category of algebraic vector bundles over X, is a category with exact sequences. Since a vector bundle is determined by its sections over open sets, Vect X is the same as the category of finitely generated locally free Ox-modules, where 0~ is the sheaf of germs of regular (algebraic) functions over X. As such, it may be identified with an additive subcategory of the abelian category of Ox-modules. A major difference between this example and Example (6) is that short exact sequences of algebraic vector bundles, unlike short exact sequences of topological vector bundles, do not necessarily split. This is due to the fact that in the algebraic setting, one does not have partitions of unity, and thus it is not possible to mimic the proof of Theorem 1.6.3.

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A related (usually slightly larger) category with exact sequences is CohShX, the category of coherent sheaves over X; this is the category of finitely generated Ox-modules with resolutions by modules from Vect X. One can show that CohShX is an abelian category. Under suitable regularity assumptions (e.g., X a nonsingular variety), resolutions of coherent sheaves by locally free sheaves will have finite length, and the relationship between the two categories Vect X and CohShX is then the same as between Proj R and R-Modf, when R is a left regular ring. 3.1.4. Proposition. Let R be a ring and let o+M+MZMz-,O be a short exact sequence of R-modules. If MI and MZ have resolutions of length n by modules in Proj R (of the form (3.1.3)), then so does M. In particular, R-Modf,,, (as defined in Example 3.1.2(4)) is a category with exact sequences. Proof. Choose resolutions 0 -_)

Pij'

_p

_+$ __)

.

.

.

_pt,

& Mj

+ 0,

j = 1,2.

By projectivity of Pi2), there is a map SF’ : Pg(‘) + M with ,6oSr’ = -$‘I. Then using r!‘, we can extend this to a surjection b. : PO = P,(l) @ PC2) 0 + M since two elements of M with the same image in M2 differ by an element of (Y(M~). Then we have a short exact sequence 0 -+ ker #’ + ker 60 -+ ker 7:’ + 0 and we can repeat the process to get a surjection 61 : PI = Pi” CD PC2)1 + ker 61. Continuing, we eventually get a resolution of M by the Pj = Pj(“@P,!‘).

Cl

3.1.5. Proposition. Let A be an abelian category, for instance the category of R-modules for some ring R, and let 0+M$M~M2--)0 be a short exact sequence in A. Assume Ml is of length n1 and M2 is of length n2 in the sense of Example 3.1.2(5). Then M is of length n1 + n2. In particular, dH is a category with exact sequences. Proof. The proof is by induction on n2 = l(M2). If this is 0, the result is obvious, and if it’s 1, this is true by definition. Otherwise, assume the result for smaller values of C(M2) and choose an exact sequence

1. Ko and K1 of categories, Go and GI of rings

113

with N of length n2 - 1 and S simple (by definition of [(M2) ). Let M’ = P-‘(N). By m’ UC d tive hypothesis, l(A4’) = ni + 722 - 1, and we have a short exact sequence O-+M’+M-+S+O, so M is of length ni + 722.

0

Now that we have a reasonable number of examples to work with, we are ready to define Ko and K1 for categories and Gc and Gi for rings. 3.1.6. Definition. Let P be a category with exact sequences with small skeleton Po. We define Ko(P) to be the free abelian group on Obj Pe, modulo the following relations: O-(i) [P] = [P’] if there is an isomorphism P -% P’ in P. 0-(ii) [P] = [Pi] + [Ps] if there is a short exact sequence

in P. Here [P] denotes the element of Ko(P) corresponding to P E Obj Po, and O-(i) is really the special case of O-(ii) with PI = 0. Note also that since every P E Obj P is isomorphic to an object of PO, the notation [P] makes sense (by O-(i)) for any object of P. We define Kl(P) to be the free abelian group on pairs (P, a), where P E Obj Ps and (Y E Aut P, modulo the following relations: I-(i) [(P, o)] + [(P, P)I = [(P, @)I. 1-(ii) If there is a commutative diagram in P with exact rows

I

a1

1

a

1

w

1

I

0 - PI L P n P2 - 0 , where cz E Aut P, ~1 E Aut PI, and cq E Aut Ps, then

[(P, a>1 = [(Pl, 41 + [P2, a2119 If R is a ring (with unit), we define Go(R) = Kc(R-Modf,), Gl(R) = Ki(R-Modfg). This definition is justified by the fact that in the case of Example 3.1.2(2), it gives us back our old definitions of Ko and K1 for rings. 3.1.7. Theorem. If R is a ring and Proj R is the category of finitely generated projective modules over R, then Ko(R) may be identified naturally w i t h Kc(Proj R), and Kl(R) may be identified naturally with

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3. KO and K1 of Categories, Negative K-Theory

K1 (Proj R). In particular, if R is a division ring, then since Proj R = R-Modf,, Go(R) = &(R) 2 Z and GI(R) = KI(R) Z Rzb.

Proof. (1) By their definitions, F&(R) and &(Proj R) are both abelian groups with one generator [P] for each isomorphism class of finitely generated projective modules over R. In Ko( R), [P] + [Q] is defined to be [P $ Q], whereas in &(Proj R), by relation 0-(ii), [P] + [Q] is given by [N] for any finitely generated projective module N for which there exists a short exact sequence O-,P+N+Q+O. Since N = P @ Q clearly has this property, the addition operations in the two groups coincide. Finally, we need to see that any relation satisfied in one group is satisfied in the other. By the definition of the Grothendieck group (cf. Theorem 1.1.3), Ko(R) is the free group on the generators [P] modulo the relations [P] = [P’] if P g P’, [P] + [Q] = [P @ Q]. These relations are satisfied in &(Proj R), so we only need check that relation 0-(ii) of Definition 3.1.6 is satisfied in Ko(R). But if

is a short exact sequence in Proj R, this sequence must split since Pz is projective, and thus P E PI @ P2, so that [p] = [PI @ p2] = [PI] +

[Pz]

in &(R),

as required. (2) If A E GL(n, R), then A defines an automorphism CY E Aut(R”), so let us define a map ‘p : KI (R) + K~(Proj R) by [A] H [(R”, a)]. To show this is well defined, suppose A’ E GL(n’, R) defines cr’ E Aut(R”‘). Recall that [A] = [A’] in Kl(R) if and only if there is some N 2 n, n’, such that

(A @ lN+) z (A’ @ 1~~~‘)

mod E(N, R).

But first of all,

[CR”, a)1 = [(RN,

a

as lR~-n)]

and

[(R”‘, a’)] = [(RN, a’ @ lRi.--n~)]

in Kl(Proj R) by relation 1-(ii) of Definition 3.1.6. Secondly, if B E GL(N, R) defines p E Aut(RN) and C E GL(N, R) defines y E Aut(RN), then BC E GL(N, R) defines r/? E Aut(RN) (we are letting matrices act on the right), and thus (by l-(i) of Definition 3.1.6)

p( [B] . [Cl) = ‘~([Bcl) =

[(RN, rP)l = [(RN r>l + [(RN > @)I, 7

which is the same as cp([B]) + cp([C]). So to complete the proof that p is well defined, we need only show that cp([C]) = 1 if C E E(N, R). It suffices

1. KO and K1 of categories, Go and G1 of rings

115

to prove this with C = eij(a), a E R. But note that there is a commutative diagram with exact rows

0 -

RN-' -k RN ?F R - 0,

where L is the obvious map from RN-’ to the vectors in RN with i-th coordinate 0, and 7r is projection onto the i-th coordinate, so that by relation 1-(ii) of Definition 3.1.6, we have [(RN, eij(o))] = [(RN-‘, URN--I)] + [(R, IR)] = [(RN, ~RN)I. Thus cp is well defined, and the proof has shown at the same time that it is a homomorphism. Now let us show that cp : K1 (R) -+ K~(Proj R) is an isomorphism. (Note by the way that we are writing Kl(R) multiplicatively and Kl(Proj R) additively.) To show ‘p is surjective, it suffices to observe that if P E ObjProj R and 01 E Aut P, then there must be (by Theorem 1.1.2) some Q E Obj Proj R and N E N with P @ Q % RN. Using relation 1-(ii) of Definition 3.1.6, we have

[(p, a)] + I(&,

lQ)]

= [(p@ Q, 0 @ lQ>l,

which, since P @ Q % RN, lies in the image under cp of GL(N, R). But [(Q, lo)] is the iden ti ty element of Kr (Proj R), so this shows [(P, a)] lies in the image of ‘p. So it remains only to show injectivity. Suppose cp([C]) = 0 for some C E GL(n, R). This means that if y is the corresponding automorphism of R”, then [(R”, r)] 1ies in the subgroup of the free abelian group on all pairs [(P, a)], P E Obj Proj R and (Y E Aut P, generated by the relations

KPY a)1 + KC @)I - KC 417 [(PY Q)I - [(Pl, %>I - KP27 a211

associated to l-(i) and 1-(ii) of Definition 3.1.6. But these relations can all be rewritten as linear combinations of the relations

[(p,

a>] - [(p @ Q, QI @ IQ>]

whenever P @ Q E R”, together with the relations associated to l-(i) and 1-(ii) with all modules not just projective but free. So we can suppose [(R”, r)] lies in the subgroup generated by relations associated to finitely generated free modules. Since we may take our finitely generated free modules to run over the set {R” : n E RI}, we may identify each automorphism of a free module with

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3. Ko and K1 of Categories, Negative K-Theory

the corresponding matrix, and we may suppose that in the free abelian group F on generators [A, j], with A E GL(j, R), j E N, [C, n] lies in the subgroup generated by the relations 1-(i’) [A, A + P, A - PA, A corresponding to the relations associated to l-(i), and by the relations l-(ii’) [A, j + ICI - [Ai, jl - L-42, 4 attached to diagrams 0 - Rj 2 @+k “--,R”-0 0 0 -

1

4

Rj 2

1

A

Rj+k

1 A

A2

I/

R” - 0 ,

corresponding to the relations associated to 1-(ii). We may further rewrite the relations of type l-(ii’) as linear combinations of those of two sorts: relations 1-(ii/)-a

[A, A - PAB-l, f (corresponding to the case k = 0 above), allowing for arbitrary changes of basis, and relations 1-(ii/)-b corresponding to the case where the injection Rj + Rj+k is the standard one given by the first j coordinates. The quotient of the free abelian group F by the subgroup generated by relations 1-(i’) and 1-(ii/)-a is clearly the direct sum Bj GL(j, R)ab. Dividing by the subgroup generated by the relations 1-(ii’)-b then gives l%GL(j, R)ab = GL(R),b = Kl(R), divided by the additional relation that

[(: 12)] = [(Agi j2)]*

However, this relation is already satisfied in Kl(R), so [C] = 1 E Kl(R) and ‘p is an isomorphism. 0 Let us now examine the meaning of Definition 3.1.6 for the other Examples 3.1.2. When X is a compact Hausdorff space, it is obvious that Ko(Vect X) is the Grothendieck group of the semigroup of isomorphism classes of vector bundles over X, and may be identified with K’(X). Kl(Vect X) is a less familiar object, but since Vect X E Proj R with R = C(X) by Theorem 1.6.3, this is the same as Ki(C(X)). It turns out (see Exercise 3.1.23 below) that there are exact sequences of abelian groups 0 + C”(X) x K1(VectwX) + KO-‘( X) -+ 0, 0 + C(X, Z) z c@(X) x Kl(Vect@X) + KU-l(X) -+ 0. The example of finite-dimensional representations of a topological group G is a special case of Example 3.1.2(5), so we turn to this sort of situation next. The following result was pointed out by Grothendieck in his earliest investigations of K-theory.

1. KO and KI of categories, Go and G1 of rings

117

3.1.8. Theorem (“D evissage”) . Let A be an abelian category in which every simple object is isomorphic to one and only one element of some set S & Obj A. Then (1) Kc(d~) is canonically isomorphic to the free abelian group on the set S. (2) Ki(da) is canonically isomorphic to eMES Ki(End M). (Since for M E S, End M is a division ring, we have Ki(End M) g (End M),X, = (Aut M)ab by Corollary 2.2.6.) Proof. (1) Clearly there is a homomorphism cp from the indicated free abelian group F to G = Ks(dH), defined by sending a generator [Ml, M E S, to the corresponding generator of G. To define an inverse $ to this homomorphism, if M E Obj A is of finite length, map [M] E G to Ci[Mi] E F, where the Mi E S are the composition factors of M (repeated according to their multiplicities), which are well defined by the JordanHGlder Theorem. This gives a well-defined map on G since if

is a short exact sequence, the composition factors of M (counting multiplicities) are just the union of the composition factors of M’ and the composition factors of M”. We have II, o cp = 1~ by the construction. To prove that cp o 1c, = lo, we show cp o $([M]) = [M] for M E Objdfl by induction on f?(M). If l(M) 5 1, this is obvious, so assume the result for M’ with l(M’) < l(M), and choose a short exact sequence

with M” E S. By inductive hypothesis, cp o +([M’]) = [M’] and ‘p o $([M”] ) = [M”] . But [M] = [M’] + [M”] , so ‘p o @([Ml) = [Ml, and this completes the inductive step. (2) Let 4,. denote the category of semisimple objects in A, i.e., the finite direct sums of simple objects. We will define an isomorphism ‘p : K1 (da) -+ eNeS K1 (End N) as follows. Given M E Obj Aa and a E Aut M, note that the largest semisimple subobject Ml of M (this is usually called the socle of M, denoted sot M) exists and must be non-zero, and is necessarily a-invariant. So there is an o-invariant canonical finite filtration of M with composition factors Mi in Obj A.. (Take the cokernel of sot M H M, take its socle, and keep iterating the construction as many times as necessary.) Let ai be the automorphism of the composition factor Mi induced by o. By relation 1-(ii) of Definition 3.1.6, we have [(M, cr)] = Ci[(Mi, q)]. Now each Mi is isomorphic to a direct sum of simple objects N E S with certain multiplicities ny, and End Mi 2 nNEs M+ (End N). So oi may be viewed as an element of &es GL(n”, End N) (this is really a finite product), and thus defines an element [cyi] of eNES K1 (End N). We let cp ([(M, cx)]) = Ci[cxi]. This defi nes a homomorphism from the free abelian group on the pairs (M, o) to eNES K1 (End N), and since it is clearly compatible with

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3. KO and K1 of Categories, Negative K-Theory

relations l-(i) and 1-(ii) of Definition 3.1.6, it passes to a homomorphism ‘p : Kl(dd + GAvEs K1 (End N). Furthermore, cp is clearly surjective, since if N E 5’ and a E Aut N, cp([(N, a)]) = [u] E (Aut N)& = Ki(EndN), and thus the image of cp contains a set of generators for eNEs Ki(EndN). It remains only to show that cp is injective. For this it is enough to note that the proof of surjectivity of cp in fact gives a construction of an inverse, namely, ifNi,..., Nk are distinct elements of S and ai E Aut Ni,

This is well defined since replacing each oi by a conjugate element of Aut Ni does not change the Ki-class on the right, and 1c, obviously gives a right inverse to cp. To see that $ gives a left inverse to cp, note that with (M, o) as above,

$0 p ([(M, a)]) =

C$([ai]) = i

@N,, @detai K

i

i

I

7

which agrees with (M, o) by the proof of the fact that Ki(Proj End Ni) g Kr(End Ni) (Theorem 3.1.7). 0 The next theorem, also due to Grothendieck, applies to our other main classes of examples, and relates R-Mod+, to Proj R and (in the regular case) CohShX to Vect X. The version in which we state it, taken from [BassHellerSwan], is probably not as general as possible, but will be adequate for our purposes. First we need a simple observation about the functoriality of Ko and K1, a simple lemma about the “Euler-Poincare principle” of $1.7, and a lemma about “resolutions” in a category with exact sequences. 3.1.9. Proposition. Suppose P and M are categories with exact sequences, and F : P -+ M is an exact functor, i.e., a functor sending short exact sequences to short exact sequences. Then F induces homomorphisms F, : Ko(P) + Ko(M) and F, : K1(P) -+ Kl(M). In fact, KO and KI are functors from the category of all categories with exact sequences and exact functors to the category of abelian groups. Proof. This is immediate from the fact that F sends relations O-(i), O(ii), l-(i), and 1-(ii) for P t o corresponding relations for M. 0 3.1.10. Lemma. Let M be a category with exact sequences contained in some abelian category A, and assume that if

is a short exact sequence in A and Mz, MS E Obj M, then Ml E Obj M. (In other words, M contains the kernel of each of its morphisms which is an epimorphism in A.) Then for any exact sequence

1. Ko and K1 of categories, Go and G1 of rings

119

the Euler characteristic Cj (-l)j [Mj] vanishes in Ko(M). Proof. This is true by O-(i) of Definition 3.1.6 if n = 1, and by 0-(ii) of Definition 3.1.6 if n = 2. So let n > 3 and assume by induction on n that the Lemma is true for exact sequences of shorter length. By the assumption on M, the kernel K of It41 + Me lies in M, so we can split the given exact sequence into the shorter exact sequences

0 + M, --+ . . . +M2+K+0.

By 0-(ii), [K] + [MO] = [MI], and by inductive hypothesis, [K] - &-I)j[M3] = 0. j=2

Combining these two equations gives ~~=e(-l)j [Mj] = 0.

0

3.1.11. Lemma. Suppose M and P are categories with exact sequences, both contained in the same abefian category A, and with P a full subcategory of M. Also assume: (1) that for each object M E ObjM, there is a finite resolution by objects of P, i.e., an exact sequence (3.1.3) in M of finite length with Pj E Obj P; (2) that if 0 --) MI + M2 --) MS + 0 is a short exact sequence in A and M2, M3 E Obj M (resp., Obj P), then MI E Obj M (resp., Obj P). (In other words, M and P each contain the kernels of each of their morphisms which are epimorphisms in A.) Then if M’ -% M is a morphism in M and

is a resolution of M by objects of P, one can complete these to a commuting diagram in M 0 + ... -

I/ 0 + . . . -_)

P'n+l -_) P,', + . . . - PA 5 M’ - 0 1 0

4 4 II --I -_) P, --_, . . . - PO -2 M --+O

whose rows are finite resolutions by objects of P. Proof. Note that PO @ M’ (E, M is an epimorphism since PO 5 M is, hence by hypothesis (2) on M it has a kernel B H PO @ M’ in M. (This

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3. KO and K1 of Categories, Negative K-Theory

is the “pull-back” of E and o.) Hence by hypothesis (1) on M, there is an epimorphism Ph -H B. Composing with the maps B + PO and B -+ M’ we get a commuting diagram

a0

1

B

1

I

J PO A M - 0. The remaining CY~, j > 1, are constructed by induction on j. Suppose j > 1 and ok has been constructed for 0 < Ic < j. By hypothesis (2) on M, Ph 5 M’ and PO -f-+ M have kernels 2; and 20 in M. Then Pi -+ Z(, and PI + 20 are epimorphisms, and also have kernels in M. Iterating the argument, we see we have Zi_i, Zj-i E Obj M and a commutative diagram with exact rows

Now we just repeat the above construction to fill in the commutative diagram

1 resaj-1 1

Pj’ - Z!3-l

aj

P j

-

-

Zj-1 -

0

I

0 .

This completes the inductive step, so the induction gives us a commuting diagram with exact rows 0 -

z:, -

E’

PA - ... - 6 -M’------+O

a1 Qn 1 a0 1 1 o - o - pn _, . . . - PO E M I/

I

- 0.

We complete the diagram by using hypothesis (1) to get a finite resolution of 2; by objects of P. 0 3.1.12. Corollary. Under the same hypotheses as Lemma 3.1.11, if M E Obj M and P. -% M, P: 5 M are two different finite resolutions of M by objects of P, then Cj(-l)j[Pj]p = Cj(-l)j[PJp in Ko(P).

1. I(0 and K1 of categories, Go and G1 of rings Proof. Apply the

121

Lemma to complete the diagram

P;

+...--t

P{

2

M

-0

(a?&, a;, (W, a;, 1 i1 Al II 0 4 P, @ PA + ... + PO @PA =L M@M - 0 , where A is the diagonal map. Consider cr. and o: as chain maps of bounded chain complexes in P: oy. : P:’ -+ P.

a n d CY~:P~--+P:.

Note that we have cut off the M’s at the end, so that P:‘, P., and P: are only chain complexes, not exact sequences. Each one is acyclic except at the 0-th slot and has non-vanishing homology Hs = M. Since cr, and CX: are isomorphisms on homology because of the commutative diagram above, the mapping cones C, and C,, are acyclic (recall Theorem 1.7.7; we are in a general abelian category rather than the category of modules over a ring, but otherwise the proof is the same). So by Lemma 3.1.10 (applied in the category P), together with the definition of the mapping cone, we have 9 = x(G) = x(P.) - x(X),

9 = x(GO = x(PL) - x(P:‘)

in &(P). The result follows immediately.

Cl

3.1.13. Theorem (“R esolution theorem”) . Suppose M and P are categories with exact sequences, both contained in the same abelian cat+ gory A, and with P a full subcategory of M. Also assume:

(1) (2)

that for each object M E Obj M, there is a finite resolution by objects of P, i.e., an exact sequence (3.1.3) in M of finite length with Pj E Obj P; that if is a short exact sequence in A and M2, MS E Obj M (resp., Obj P), then MI E Obj M (resp., Obj P). (In other words, M and P each contain the kernels of each of their morphisms which are epimorphisms in A.)

Then the inclusion functor P v M induces an isomorphism on Ko. Proof. If a category with exact sequences P is a full subcategory of a category with exact sequences M, then the inclusion functor L : P - M is exact, so it induces a map L* on KO and KI by Proposition 3.1.9. Using the idea of the Euler characteristic from $1.7, we construct an inverse map

~PO : Ko(M) + Ko(P) by PO : WIM H Cj(-l>j[Pj]P if 0 + P, + . . .

-+P,+M+O

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3. KO and KI of Categories, Negative K-Theory

is exact in M, with Pj E Obj P. This is well defined, i.e., independent of the choice of resolution, by Corollary 3.1.12. By Lemma 3.1.10, [M]M = Cj(-l)j[Pj]~ in Ko(M), so L* o yx,([M]~) = [M]M. On the other hand it is clear that cpo o h,([P]p) = cpo([P]~) = [P]p for P E Obj P, so L* is an isomorphism on Ko. 0 Next we have the analogue of Theorem 3.1.13 for K1, again following [BassHellerSwan]. Note that the hypotheses are a bit stronger than those of 3.1.13, though they will still be satisfied for all cases of interest. 3.1.14. Theorem (“R esolution theorem for ICI”) . Suppose M and P are categories with exact sequences, both contained in the same abelian category A, and with P a full subcategory of M. Also assume: (1) that for each object M E Obj M, there is an epimorphism P + M in A with P an object of P, such that every endomorphism of M lifts to an endomorphism of P; (2) that if ...

4-l

+P$+P,_1 - . . . + PO + M -i 0

is exact in M with Pj E Obj P, then kerd, E Obj P for some (sufficiently large) n; (3) that if 0 --+ Ml + M2 + M3 --) 0 is a short exact sequence in A and Mz, MS E Obj M (resp., Obj P), then Ml E Obj M (resp., Obj P). (In other words, M and P each contain the kernels of each of their morphisms which are epimorphisms in d.) Then the inclusion functor P -+ M induces an isomorphism on K1. Proof. First we want to show that every automorphism of an object of M lifts to an automorphism of some finite resolution of M by objects from P. Then we will be able to apply the same sort of reasoning as in the proof of Theorem 3.1.13. So let M E ObjM, (Y E Aut M. Using (l), choose P ++ M w i t h P E Obj P so that every endomorphism of M lifts to an endomorphism of P. Then consider Q @ 6’ E Aut(M @ M). By Lemma 1.5.4 (the same argument works in a general abelian category), this factors as a product of “elementary” automorphisms of the form

with p, y E End M. Lifting p and y to endomorphisms of P shows that Q $ ay-’ lifts to an automorphism of P @ P. Then (P + M) @ (P + 0) gives us the first step of our desired resolution of M. The kernel of this map P @ P + M must be an object of M by hypothesis (3), so we can

1. KQ and K1 of categories, Go and Gl of rings

123

repeat the same construction over and over to get a (potentially infinite) resolution of M by objects of P so that cr lifts to an automorphism of the resolution. Then we can cut off the resolution at some finite stage by hypothesis (2). The rest of the proof is as in Theorem 3.1.13. We construct an inverse (pl

to b* : Kl(P)

-+ KlW) by mapping

[CM,

(~11 +-+ Cj(-l)j[(pj, q)l,

where a. is an automorphism of the finite resolution P. of M lifting CY. To show this is well defined (and independent of the choice of resolution), we use Corollary 3.1.12 applied to the categories of pairs (M, a), (Y E Aut M, M E Obj M (resp., P), where the morphisms are commutative diagrams M a

1

M

f M’ a’

f

1

M'.

It is easy to see that the hypotheses of Corollary 3.1.12 apply to this situation, and we finish the proof as in Theorem 3.1.13. 0 In order to apply Theorems 3.1.13 and 3.1.14 to the context of Rmodules, we need as a preliminary a familiar fact from homological algebra. (See, for instance, [CartanEilenberg, Proposition VI.2.11.) 3.1.15. Lemma. Let R be any ring (with unit) and let M be an Rmodule. Then the following are equivalent: (a) M has a projective resolution of length n. (b) For any R-module N, Extl+‘(M, N) = 0. (c) The functor N + Extz(M, N) is right exact. (d) For any projective resolution dntl . . . + pm+1 -

4-l . . . + PO 3 M --f 0, pn ++ P,__l __*

imd, = kerd,_l is projective, and hence the resolution can be shortened to 4-l

O+imd,+P,_~-... + PO -% M + 0. Proof. (a) + (b). Suppose 0 + P,, -+ . . . -+ PO --+ M + 0 is a projective resolution of M. Then by definition, Exti(M, N) is the j-th homology module group of the complex HOmR

(PO, N) + . . . + HOmR (P,, N) + 0 + . . . ,

so clearly Extg+’ (M, IV) = 0. (b) + (c). Assume Exti+l (M, IV) = 0. Given a short exact sequence 0 + Nl -+ N2 + N3 + 0,

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3. KO and K1 of Categories, Negative K-Theory

there is an associated long exact sequence Ext;(M, Ni) + Ext;(M, Ns) + Ext;(M, Ns) + Ext”R+l(M, Ni) = 0, and thus the functor N + Extk(M, N) is right exact. (c) + (d). To check the projectivity of imd,, we need to show that given a short exact sequence

and a homomorphism (Y : im d, + Na, cr factors through Ns. In other words, we need to show that the natural map HomR (imd,, Nz) + HomR (imd,, Ns) is surjective, i.e., that the functor N + HomR (imd,, N) is right exact. This is immediate from (c) if n = 0, so assume n > 0. From the short exact sequences 0 ---f im dj+l = ker dj + Pj + imdj --+ 0,

j > 0,

{ O-+imdi=kerdc+Ps-+M-+O, we obtain exact sequences 0 = ExtkV1(Ps, Nj) + Extk-‘(imdi, Nj) --+ Ext;(M, Nj) + Extn,(Ps, Nj) = 0, Ext;-j(P+i, Nj) + Ext;-j (imdj, Nj) + Extl-j+‘(imdj-1, Nj) + Ext;--j+’ (P+ NJ = 0, 0 < j 5 n. These yield isomorphisms Extl;;(M, Nj) 2 Extk-‘(imdi, Nj) 2 . . . E Ext]R(imd,_i, Nj) and thus an exact sequence HomR (Pn_l, Nj) + HomR (imd,, Nj) + Extg(M, Nj) + 0. Assuming (c) and using projectivity of P,_l, we obtain a commutative diagram with exact rows and columns HomR (P,_l, Nz) + HOmR(imd,, N2) -+ Extz(M, Ns) +

&mR (Pn_l, N3) -+ HOmR (imA, N3) + ExtE(“~

1

0

1

0

and surjectivity of HomR (im d,, Nz) -+ HomR (im d,, diagram chase. (d) + (a) is trivial. •I

0

N3) - o

/I 0,

N3) follows from a

1. KO and K1 of categories, Go and GI of rings

125

3.1.16. Corollary (Grothendieck). Let R be a left Noetherian ring (with unit). Then the natural map Kj (R) + Kj (R-Modf,,) (induced by the inclusion Proj R q R-Mod+, together with Theorem 3.1.7) is an isomorphism for j = 0, 1. In particular, if R is left regular, then the natural map Kj (R) -+ Gj (R) (induced by the inclusion Proj R q R-Modf, together with Theorem 3.1.7) is an isomorphism for j = 0, 1. Proof. We need only check the hypotheses of Theorems 3.1.13 and 3.1.14. It is clear from the definition that every R-module in R-Mod@, has a finite resolution by finitely generated projective modules. Furthermore, given an epimorphism P + M with P surjective, and given an endomorphism (Y of M, we can fill in the diagram P

-

M a

P

-

1

M

-

0

-

O

I

by projectivity of P to get a lifting 6 of cz to P, which checks hypothesis (1) of Theorem 3.1.14. Next, every epimorphism in Proj R splits, hence has kernel which is a direct summand in a projective module, hence has a kernel in Proj R. Hypothesis (2) of Theorem 3.1.14 holds by the implication (a) + (d) of Lemma 3.1.15. To finish the proof, we need only show that if M and M’ have finite projective resolutions of finite type and if M -?+ M’ is an epimorphism, then kercr also has a finite projective resolution of finite type. First of all, if M and M’ each have projective resolutions of length n, then by the implication (a) + (b) of Lemma 3.1.15, ExtE+‘(M, N) = Extz+l(M’, N) = 0 for any R-module N. By the long exact sequence associated to the short exact sequence O+kera+M+M’+O, 0 = Ext;+l(M, N) -+ Exti+‘(ker (Y, N) + Ext:+2 (M’, N) = 0 is exact, so Exti+‘(ker o!, N) = 0 and ker a has a projective resolution of length n by the implication (b) + (a) of L emma 3.1.15. If R is left Noetherian, it is immediate that ker CY in fact has a projective resolution of finite type, since we can start with any resolution of ker CY by finitely generated free modules (such a resolution exists, since any submodule of a finitely generated module is finitely generated) and truncate it using the implication (a) =+ (d) of Lemma 3.1.15. Cl Remark. The same sort of reasoning shows that if X is a non-singular projective algebraic variety, then the natural map Kj (Vect X) * Kj (CohSh X) is an isomorphism for j = 0, 1. We omit the proof since setting up the necessary machinery requires knowledge of too much algebraic geometry.

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3. KO and K1 of Categories, Negative K-Theory

Now we are prepared to explain the idea of Grothendieck’s original motivation for studying the K-theory of categories, namely, for use in studying the Riemann-Roth problem. Grothendieck in fact substantially generalized both this problem and the form of its solution, but for simplicity we will restrict attention here to the classical situation. For readers who are unfamiliar with it, we begin with a quick review of the terminology of sheaf theory. If X is a topological space, a presheaf 3 (say of R-modules) over X is a contravariant functor from the category of open sets of X (and morphisms given by inclusions) to the category of R-modules. The notation I’(U, 3) is also used for 3(U), and we refer to this module as the sections of 3 over U. A sheaf 3 of R-modules is a special kind of presheaf: one which also satisfies the gluing condition, that if {Uj} is a collection of open subsets of X, the restriction map

is a bijection. Typical examples of sheaves are the sheaf of germs of continuous R-valued functions, whose module of sections over U is C”(U), and the structure sheaf 0~ of an algebraic variety, whose module of sections over a Zariski-open set U is the set of regular (algebraic) functions defined in U. These may be viewed as rational functions without poles in U. In the category of sheaves over X, the global section functor 3 y-) I’( X, 3) is left exact but not right exact. It has derived functors Hj (X, 3), with the properties that H”(X, 3) = I’(X, 3) and that a short exact sequence of sheaves 0 + 31 --+ 3s --) 3s -+ 0 gives rise to a long exact sequence ... + Hj(X, 3i) + Hj(X, 34 -+ Hj(X, 3S) -+ Hj+‘(X, 3i) + . . . . We return now to the classical Riemann-Roth problem. Let X be a nonsingular projective algebraic variety of dimension 1 over @, or for short, a nonsingular curve. X is a compact connected complex manifold of complex dimension 1 and real (or topological) dimension 2, or in other words a compact connected Riemann surface, say of genus g. (Recall that the genus is a purely topological invariant of the underlying manifold of X that doesn’t depend on the algebraic structure. It may be defined as the number of “holes” in X, or more precisely as i rank Hl(X; Z) = 3 dim@ H1(X; e).) A divisor D on X is just a formal finite Z-linear combination c njzj of points zj E X, with nj E Z. The divisors D are in bijection with isomorphism classes of algebraic line bundles over X via the map C njxj = D H Co, where Ice is the line bundle whose (algebraic) sections over an open set U are the rational functions f over U vanishing to order at least -nj at x~j (and thus regular at points x where n, 5 0). By convention, we say f vanishes to order 0 at z if f(x) E Cx, and f vanishes to order -lc at x, k > 0, if f has a pole of order k at x. We make the usual identification of

1. KO and KI of categories, Go and G1 of rings

127

a line bundle with the sheaf of its (algebraic) sections. This is a locally free Ox-module of rank 1; in general, locally free Ox-modules of finite rank correspond to algebraic vector bundles over X. Note that LE1 = L-D, in the sense that Lo @IX c-0 = OX, with @x the tensor product for sheaves (computed pointwise over X). The classical Riemann-Roth problem was to compute the dimension a(D) of the space I’(X; LO) of global (algebraic) sections of LD, for any divisor D. We may think of this dimension as a Betti number for sheaf cohomology, namely, as dimHO(X, LO). For instance, if D = 0, LD = 0~ and l(D) = 1 (since any rational function on X without poles must be constant by compactness and the maximum principle for analytic functions). The Riemann-Roth Theorem (see for instance [Hartshorne, Ch. IV, $11) asserts that (3.1.17)

C(D) - l(K - D) = deg(D) + 1 - g,

where LCK is the canonical sheaf (the sheaf of algebraic l-forms f(z) dz) and the degree of a divisor is defined by deg C njxj = C nj. The formula (3.1.17) is clear if D = 0, since a(O) = 1, deg0 = 0, and e(K) = dim H”(X, LK) is the dimension of the space of algebraic l-forms on X, while by the Hodge Theorem, 2g = dimHl(X; Cc) = dim{harmonic l-forms on X} = dim ({holomorphic l-forms on X} @ {anti-holomorphic l-forms on X}) = dim H’(X, LK) + dim He(X, ,CK) = 2 dim H’(X, LK), so that e(K) = g. Let us now sketch a proof of (3.1.17) using &(CohShX), the K-theory of the category of coherent sheaves on X. Since X is non-singular and of (complex) dimension 1, any coherent sheaf F over X has a resolution of length 1 by locally free sheaves: (3.1.18)

0 + vi -+ vo + F + 0.

Furthermore, Hj(X, F) is finite-dimensional for j = 0, 1 and vanishes for j > 1. One may prove this by using the long exact sequence in sheaf cohomology associated to (3.1.18) to reduce to the case of a vector bundle V. The finite-dimensionality of Ho comes from compactness of X and Montel’s Theorem (which says that the space of holomorphic sections of V over X must be locally compact, hence finite-dimensional). The Serre duality theorem says dim H’ (X, V) = dim H”(X, 0 @x LK), where 3 is the “dual” bundle to V (in the case of a line bundle this is just V-l). Hence we have finite-dimensionality of H1 as well. The vanishing of the

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3. KO and K1 of Categories, Negative K-Theory

cohomology past the (complex) dimension follows, for instance, from Dolbeault’s Theorem, which identifies Hj(X, V) with the j-th cohomology of the complex of antiholomorphic differential forms with values in V. Thus for 3 a coherent sheaf over X, the Euler characteristic m

x(3) =

c Hj(X, 3)

j=O

is well defined and given just by dim H’( X, F) -dim H1 (X, 3). For a line bundle ,C, Serre duality gives that X(,C) = dim H’( X, L) - dim H’( X, L-l 631~ LK). In particular, e(o) - !(K - 0) = dim H’( X, LO) - dim H’( X, LK 63~ LE1) = I, so the Riemann-Roth Theorem amounts to the statement that (3.1.19)

x(LD> - X(0x) = deg D.

To prove this, note that by the “Euler-Poincare Principle” (cf. Proposition 1.7.10), for any short exact sequence of coherent sheaves o-+31 -32+3s+o, we have additivity of the Euler characteristic: X(32) = X(31) + X(33). (One may see this by taking the corresponding long exact sequence in sheaf cohomology and applying Lemma 3.1.10 in the category of finitedimensional vector spaces over Cc.) It follows that the map 3 I--+ X(3) preserves relation 0-(ii) of Definition 3.1.6 and thus passes to a homomorphism X : I& (CohSh X) -+ Z. It will suffice for us to get a better understanding of this homomorphism. The trick (which was the key contribution of Grothendieck) is that even though we were initially only interested in X(3) in the case of line bundles, it pays to study X in the larger category CohShX where we have more exact sequences and thus more non-trivial relations to help us. Let x be a point of X and let D be any divisor. There is a natural monomorphism LD -+ LD+I coming from the fact that every section of LD is also a section of Lo+. This map is an isomorphism away from 2, so the quotient sheaf S, is a coherent sheaf supported only at x. Furthermore, if n, is the coefficient of x in D, then for U small enough, I(u, LD+,)/I(& LO) iS spanned by ZH

%+I 1 z - x

( >

1. Ko and K1 of categories, Go and G1 of rings

129

in local coordinates, and hence dimI?(U, &) = 1 if x E U. From this one can see that dimHO(X,&) = 1, dimH’(X, &) = 0, so X(&J = 1. Then from the exact sequence

we obtain X(LD+,) = X(cD)+l. Reversing the roles of D and D+x, we get X(LD_) = X(LD) - 1. So if D = cnjxj,, we get X(LD) = X(&) + xn.j, which is (3.1.19).

3.1.20. Exercise. Let R be a PID, for instance Z. Show that if M is a finitely generated torsion R-module, then [M] = 0 in Go(R) 2 Ko(R) (E Z). Is this necessarily true if R is only a Dedekind domain? 3.1.21. Exercise. Show that the analogues of Theorem 1.2.4, Exercise 1.2.8, Exercise 2.1.6 and Exercise 2.1.7 hold for Go and Gi. In other words, show that Gj is Moritainvariant and that Gj (R x S) %J Gj (R) @ Gj (S), for j = 0, 1. 3.1.22. Exercise. Consider the categories Repz of finite-dimensional complex representations of Z, which may be identified with the category of pairs (V, T), where V is a finite-dimensional complex vector space and T E Aut V is the image of the generator, and the category RepN of finitedimensional complex representations of the monoid N, which may be identified with the category of pairs (V, T), where V is a finite-dimensional complex vector space and T E EndV. Determine the simple objects in these categories and use Theorem 3.1.8 to compute Kc and Kr for each category. 3.1.23. Exercise [Milnor, $71. Let X be a compact Hausdorff space, and recall that Kj(VectFX) s Kj(R) with R = C"(X) by Theorem 1.6.3, for F = R or C, j = 0, 1. Show that there are exact sequences of abelian groups 0 --f C”( X) x Kl(VectRX) + KO-l(X) + 0, 0 -+ C(X, Z) -% c@(X) 3 Kl(Vect@X) + KU-l(X) + 0.

130

3. KO and K1 of Categories, Negative K-Theory

Here is a sketch of how to proceed. Recall from Exercise 1.6.15 that

K,-‘(x) =&f K,o(x x lit). Let S = C”(X x [0, 11) and let I be the closed ideal of functions vanishing on X x (0, 1). Then I Z C[ (X x (0, 1)) (as a Banach algebra without unit) and from the short exact sequence O+I+S-+RxR+O we obtain an exact sequence K~(S) + K1(R) @ ITI + Ko(l) + Ko(s) + Ko(R) @ f6dR)By homotopy invariance of Kc (Corollary 1.6.12), Kc(S) r Ko(R), and the map on the right may be identified with the diagonal map Ko(R) + Ko(R)@Ko(R), wh ic his injective. Furthermore, there is a splitting map from the diagonal copy of R inside R x R to S (extend a function on X to a function on X x (0, 1) which doesn’t depend on the second coordinate), so that the above exact sequence gives the exact sequence K1(S) -+ K1(VectFX) --+ K,O(X x W) + 0, where we think of Kr (VectpX) as K1 (R) @ 0 - K1 (R) @ K1 (R). Show that the image of Kr (S) -+ Kl(R) @ 0 can be identified with the classes in Kl(R) represented by matrices in GL(R) % C(X, GL(F)) which are homotopic to elementary matrices. Then show that the part of SKI(R) coming from matrices homotopic to elementary matrices is trivial, and that the classes in RX homotopic to the identity coincide with the image of the exponential map C”(X) x C(X, GL(1, IF)). (Use the idea of Lemma 1.6.6 to show that an element of C(X, SL(n, IF)) (resp., C(X, GL(1, F))) wh’ rch is close to the identity is an exponential of something in C(X, SL(n, P)) (resp., C(X, F)).) Finally, identify the kernel of the exponential map C”(X) = C(X, GL(1, IF)) for F = R, Cc. 3.1.24. Exercise. Let p be a prime number and consider the local ring R = Z/(p”) with unique maximal ideal I = (p). (1) Show that R is not left regular, by showing that R/I has a resolution by finitely generated free R-modules for which condition (d) of Lemma 3.1.15 is not satisfied for any 72.

1. KO and K1 of categories, Go and G1 of rings

131

(2) Note that R-Modfg is a category in which R/I is the unique simple object (up to isomorphism), and in which every object has finite length. Then use Theorem 3.1.8 to compute Go(R) and Gl(R). Is the natural map Kj (R) -+ Gj (R) an isomorphism for j = O? For j = l? 3.1.25. Exercise (A step toward Grothendieck’s generalized Riemann-Roth Theorem). Let X be a non-singular projective algebraic variety over C:, now of dimension n > 1. In this more general setting, a divisor D on X is defined to be a formal finite Z-linear combination C njXj of subvarieties Xj C X of codimension 1, with nj E Z. The divisors D are again in bijection with isomorphism classes of algebraic line bundles over X via the map C njXj = D H LD, where ~ZD is the line bundle whose (algebraic) sections over an open set U are the rational functions f over U vanishing to order at least -nj along Xj (and thus regular along subvarieties Y of codimension 1 for which ny 5 0). The generalized Riemann-Roth problem, solved by Grothendieck, is to give a formula relating X(Co) to x(0x), analogous to formula (3.1.19). (1) Assuming that coherent sheaves over X have finite resolutions by locally free sheaves and thus that the natural map Ko(Vect X) + Ko (CohSh X) is an isomorphism, and assuming the result of Serre that for F a coherent sheaf over X, Hj(X, F) is finite-dimensional for j 5 n and vanishes for j > n, show as in the one-dimensional case above that the map _7= +--+ X(3) preserves relation 0-(ii) of Definition 3.1.6 and thus passes to a homomorphism X : Ko(Vect X) %’ K,,(CohShX) + Z. (2) Let Y be an irreducible subvariety of X of codimension 1, taken for simplicity to be non-singular. Show as in the one-dimensional case above that there is a short exact sequence of coherent sheaves

where the quotient sheaf S D, y is a coherent sheaf supported along Y. Note in fact that SD, y E LD @x Oy, where we think of the structure sheaf of Y as being extended to a sheaf on X supported along Y. Deduce that X~(~Z-Y)-XX(OX) = -Xy(Oy), or in general that XX(cD-_Y) - Xx(LD) = -Xy(L*cD), where L : Y -+ X is the inclusion and L*LD is the pull-back of LD to a line bundle on Y. This suggests a mechanism for proving a generalized Riemann-Roth formula by induction on n. 3.1.26. Exercise (Relative K-groups for categories). Let A and B be categories with exact sequences, and let F : A -+ I3 be an exact functor, SO that it defines homomorphisms F, : Kj (A) + Kj (f?), for j = 0, 1. Define a relative group Ko(F) to be the free abelian group with generators

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[Ao, Al, a] corresponding to pairs (Ao, Al) E Obj Ax Obj A together with a morphism Q : A0 + Al in A for which F(o) : F(Ao) -+ F(A1) is an isomorphism in 8, modulo the relations that [Ao, Al, a] = 0 if CY is an isomorphism in d, and that if there is a commuting diagram of short exact sequences 0 - Ab’ - Ao -A;-0

a a’ 1 1 1 II A’i - Al -A’,---+0

a”

I 0

-

and F(a), F((Y’), F(cr”) are isomorphisms in Z?, then

[A,,, Al, a] = [A;, A;, a’] + [A;, A’,‘, a”]. Define a map Kc(F) 2 Ko(d) by

[Ao, AI, ‘Y]K~(F)

I-+ [Aoko(d) - [A1lKo(d)

and show that F, o 4 = 0. Assume further that F is what Bass calls “cofinal,” in other words that given Bi E Obj B, there is some B2 E Obj B with Bl@ B2 E F(A) for some A E Obj d, and also that one can choose the Bz so that F, : End A ---) End F(A) is surjective (this condition is similar to the first condition in Theorem 3.1.14). Show that there is an exact sequence

Kl(d)

2 KI(@

%,(F)L Ko(d)~Ko(Z?)

by imitating arguments from Theorems 1.5.5 and 2.5.4. To define the map 8, note that if B1 E ObjB and /3r E Aut(Br), then with Bs and F as above, [Bl, p] = [Bl @ Bz, pi @ lls,] in Kr(Z3) can be replaced by [F(A), ,0] with A E Objd and p E Aut F(A). Then if F, : End A + End F(A) is surjective, p lifts to an endomorphism o of A, and we can define d([F(A), p]) = [A, A, a]. One has to check that this is independent of the choice of A and (Y. Check that when _4 = Proj R, I is an ideal in R, 23 = Proj R/I, and F is induced by the quotient map R + R/I, then the hypotheses on F are satisfied and one recovers the exact sequence of Theorem 2.5.4.

2. The Grothendieck and Bass-Heller-Swan Theorems In this section, we consider the problem of computing the K-theory of a ring of polynomials or Laurent polynomials over another ring whose K-theory is already known. In the case where R = CF(X) is the ring of continuous F-valued functions on a compact Hausdorff space X (with

2. The Grothendieck and Bass-Heller-Swan Theorems

133

F = R or C), the ring of polynomials R[t] is, by the Stone-Weierstrass Theorem, a dense subring of C”(X x [0, 11) ( via specialization oft to a real number in [0, 11). The homotopy invariance theorem for topological Ktheory (Corollary 1.6.11) says that the map t H 0 induces an isomorphism Ke(C”(X x [0, 11)) + K&“(X)). Thus it is reasonable to view the map on K-theory fw[4) -+ Km induced by t H 0 as corresponding to “algebraic homotopy,” and to expect this map to be an isomorphism for suitable rings R. This turns out to be the case for R

left regular (Grothendieck’s Theorem), though homotopy invariance fails in general. The case of the Laurent polynomial ring R[t, t-‘1 is more complicated. When R = C”( X), there is a map R[t, t-‘1 q C@(X x S’) defined via specialization oft to a complex number of absolute value 1, and the image is dense by the Stone-Weierstrass Theorem. On the other hand, in topological K-theory, one has the formula

KU-j@

x

S1) E KU-j(X) @KU-j-l(X).

In complex K-theory, Bott periodicity holds and K-j only depends on j modulo 2. Thus if we specialize to j = 1, we have KU-l(X x S’) 2 KU-l(X) @KU’( X) Z KU-l(X) @ Ko(R).

Since, by Exercise 3.1.23, KU-l(X x S’) and KU-l(X) are closely related to Kl(C@(X x S1)) and to K1(R), respectively, and since R[t, t-l] is dense in Cc(X x S’), this suggests that perhaps one can expect to have K1 (R[t, t-l]) 2 K1 (R) @ Ko(R) when R is a nice enough ring. In other words, taking Laurent polynomials should correspond to “algebraic desuspension.” Again, this will turn out to be the case for R left regular (the Bass-Heller-Swan Theorem). We will also be able to study the extent to which this and algebraic homotopy invariance fail for rings which are not left regular. Finally, further study of these ideas will also lead to a definition of K-groups extending the exact sequence of Theorem 2.5.4 arbitrarily far to the right. We begin with a review of two famous theorems of Hilbert, the Basis Theorem and the Syzygy Theorem, which together imply that if R is left regular, so are R[t] and R[t, t-l]. The reader who is already familiar with these classical theorems can skip to formulas 3.2.5 and 3.2.6 and to the discussion surrounding them. 3.2.1. Theorem ( “H ilbert Basis Theorem ”) . Let R be a left Noetherian ring. Then the polynomial ring R[t] is left Noetherian. Proof. Let J be a left ideal of R[t], and consider the sets I, Ij of all leading coefficients of the polynomials in J (respectively, of the polynomials in J of degree 5 j). Since J is closed under addition and left multiplication

134

3. KO and K1 of Categories, Negative K-Theory

by elements of R, so are I and the Ij, hence I and the Ij are left ideals of R. Since R is left Noetherian, we may choose a finite set of polynomials fi(t)=a;‘P+...+a~ .,.

EJ

fn(t)=ay~+...+a; 1 whose leading coefficients a;l, . . . , a> are generators of I. Let m = maj deg fj = maxj rj. Then if f(t) = bht” + * *. + bo E J, the leading coefficient bk of f may be written in the form ~~=, cjay, Cj E R, and if Ic 2 m, then f(t) -CT=, cjt”-‘j fj(t) 1ies in J and has smaller degree than f. A simple induction thus shows that J is generated as a left ideal of R[t] by A(t), . . . , fn(t) and by the polynomials in J of degree < m. If we similarly choose successively, for j = 0,. . . ,m - 1, finitely many polynomials d, . . . ga, E J of degree 5 j whose leading coefficients generate Ij, then it is evident that the fr (t), . . . , fn(t) together with the d(t) generate J as a left ideal of R[t]. • i 3.2.2. Corollary. If R is a left Noetherian ring, then so is the Laurent polynomial ring R[t, t-l]. PTOC$. R[t, t-‘1 is a localization of R[t], and a localization of a left Noetherian ring is left Noetherian. For a more explicit proof, let J be a left ideal of R[t, t-l], and let Jo = JflR[t], which is a left ideal of R[t]. (Here we think of R[t] as a subring of R[t, t-l].) Using Theorem 3.2.1, choose finitely many generators for JO. Then these also generate J, since for f(t) E J, f(t) = t-“t”f(t), and t”f(t) E JO for n 2 0 sufficiently large. Cl 3.2.3. Theorem (“H ilbert Syzygy Theorem”) . If R is a left regular ring, then so is R[t]. firthermore, if R has (left) global dimension < n, then R[t] has (left) global dimension 5 n + 1. Proof. By the Basis Theorem, R[t] is left Noetherian. Let M be a finitely generated left R[t]-module. By Lemma 3.1.15, to show that M has a resolution of finite type by projective R[t]-modules, it will be enough to show that there exists a positive integer K such that Ext&(M, N) = 0 for all R[t]-modules N, and to prove the final statement about global dimension, we only need to show that if R has (left) global dimension < n, then K can be taken to be n + 2. By restriction, any R[t]-module can be considered to be an R-module, which comes naturally with an R-module endomorphism cp defined by left multiplication by t. So we can form the short exact sequence

0 -+ R[t] ‘8R M = R[t] @R M -i M -i 0. Suppose that for some k, Extk(M, N) = 0 for all R-modules N. In fact, if R has (left) global dimension 5 n, Ic can be taken to be n + 1. We have an exact sequence Ext&(R[t] @,R M, N) -+ Ext#M, N) + EXtk;;(R[t] ‘8R M7 N)’

2. The Grothendieck and Bass-Heller-Swan Theorems

135

Since R[t] is free as an R-module, Ext& (R[t] @R M, N) g Ext$(M, N) = 0 for j 2 Ic, so Ext&( M, N) = 0 for K = Ic + 1, as desired. To complete the proof, we only need to get around one technical point: M is assumed to be finitely generated as an R[t]-module, but may not be finitely generated as an R-module, so that in the case where R is not assumed to have finite global dimension, the definition of left regular ring doesn’t immediately tell us that M is of finite homological dimension as an R-module. The following trick for getting around this may be found in [Bass, pp. 634-6351, though part of the idea is older. Let Me be a finitely generated R-submodule of M which generates M as an R[t]-module, and let M, = PM0 + . . . + MO. Then M, is an increasing sequence of finitely generated R-submodules of M and M = 15 M,. Let Qn = {z E MO : Pa: E Mn_l} and observe that this is an increasing sequence of R-submodules of MO. Since MO is finitely generated and R is left Noetherian, there is some no such that Qn = Qn, for all n 2 no. We claim that for n > no, the homological dimension of M, is 5 d, where d is the larger of the homological dimensions of M,,, and Mm0 /M,,, -1. Indeed, this is true for n = no, and if n > no and it’s true for n - 1, we can apply the exact sequence 0 + Mn_l -+ M, + Mn/Mn_l + 0. By choice of no and the assumption that n > no, the map Mn,,/Mno-l ‘“ z” M,/M,_l has a trivial kernel and thus is an isomorphism. So for any Rmodule N, we have an exact sequence Ext; (M,o /M,o - 1, N) -+ Ext;(n/m, N) --+ ExtjR(M,+r, N), and since Exti(M,,/M,,,_i, N) = 0 and Exti(Mn_r, N) = 0, we get that Exti(M,, N) = 0. This proves the claim by induction. To complete the proof, one needs to see that the homological dimension of M is bounded by limsup horn. dim. M, 5 d. This follows from the LLl@l-sequence” 0 + 12’ Ext;--l (M,, N) --+ Ext;4(lsMnr N) + l@Ext$(M,, N) + 0 which comes from the long exact sequence in Ext associated to the short exact sequence O-+

BM, (Zn)~(~or~l-zO ?...I z,--I”-l,... 1, @ Mn n

n (“n)HC,Gl -limM, -+O. • i +

136

3. Ko and K1 of Categories, Negative K-Theory

3.2.4. Corollary. If R is a left regular ring, then so is R[t, t-l]. Furthermore, if R has (left) global dimension 5 n, then R[t, t-l] has (left) global dimension < n + 1. Proof. Again this follows from the fact that R[t, t-l] is a localization of R[t]. More explicitly, if M is a finitely generated R[t, t-‘l-module, choose generators 21, . . . , z, for M and let Ml be the finitely generated R[t]module they generate. Then M = R[t, t-l] @Rit] Ml, and since R[t, t-l] is flat over R[t], Ext$l,,,_,l(M, N) = Ext&,,&R[t, t-l] @‘~[t] MI, N) = Ext;@k N). Hence the homological dimension of M over R[t, t-‘1 is the same as that of MI over R[t]. 0 Now we’re ready to proceed with the study of the K-theory of R[t] and R[t, t-l]. Note first of all that there are split short exact sequences (3.2.5)

0 --t tR[t] + R[t] 2 R + 0,

(3.2.6)

0 + (t - l)R[t, t-l] + R[t, t-l] % R + 0,

so that the K-t,heory of R[t] or of R[t, t-l] contains that of R as a direct summand. The basic problem is to study the other summands, if any. It turns out that in this context, G-theory behaves better than K-theory, at least for rings which are left Noetherian. Hence it is worth saying something about the functoriality of G-theory under change of rings. In general, if ‘p : R --+ S is a ring homomorphism, though it induces an exact functor from Proj R -+ Proj S, cp* is usually not exact as a functor from R-Modf, to S-Modf,, hence does not induce a homomorphism Gj (R) + Gj (S). However, if S is flat over R (which is another way of saying cp* is an exact functor), in particular if S is projective as an R-module (via cp), then cp* : GJR) 4 GJS) is defined. This will be the case, for instance, when cp is the obvious injection of R into S = R[t] or R[t, t-l]. We would like, however, to have maps Gj(R[t]) --+ Gj(R) and GJR[t, t-l]) + GJR) in spite of the fact that the obvious maps R[t] = R and R[t, t-l] -% R are not flat. The device for constructing such maps, due to Grothendieck, is based on the ideas that went into the proof of the Resolution Theorem (Theorem 3.1.13). We use the fact that R has finite homological dimension over R[t] or R[t, t-l]. In fact, from the resolutions (3.2.7)

0 + R[t] 4, R[t] 2 R -+ 0, 0 + R[t, t-‘1 = R[t, t-l] = R --+ 0,

2. The Grothendieck and Bass-Heller-Swan Theorems

we see that thus that if

R

137

has homological dimension 1 over R[t] and R[t, t-l], and 0 + Mi + Mz + MS + 0

is a short exact sequence of R[t]-modules, there is a corresponding exact sequence of R-modules (3.2.8)

0 -+ Torfltl(R, Ml) -+ TorfLtl(R, Mz) -+ Torfltl(R, MS) + R @iqt] MI + R @.~[t]

M2 ---, R @R[t] M3 --+ 0,

and similarly with R[t, t-l] in place of R[t]. 3.2.9. Proposition. Let R be a left Noetherian ring. There are welldefined homomorphisms Go(R[t]) + Go(R) and Go(R[t, t-l]) + Go(R) defined by

[Ml - [R @R[t] Ml - P’o rf[tl

CR, WI

(or the same formula with R[t] replaced by R[t, t-l]). When R is left regular, these agree with the usual homomorphisms &(R[t]) + Ko(R) and Ko(R[t, t-l]) + Ko(R). Similarly, there are well-defined homomorphisms Gi (R[t]) + G1 (R) and Gl(R[t, t-l]) -+ Gl(R) defined by

[M, a] H [R @.R[t] M, 1 C3 a] - [TorfLtl(R, M), Tor(1, a)] (or the same formula with R[t] replaced by R[t, t-l]) which agree with the usual homomorphisms Kl(R[t]) -+ Kl(R) and Kl(R[t, t-l]) --+ Kl(R) when R is left regular. Proof. First consider the case of Ge and R[t]. The indicated formula gives a well-defined homomorphism for two reasons: (i) If M is finitely generated as an R[t]-module, then R@R[t] M is finitely R[tl R( ,M) is finitely generated since it may be generated, and also Tori computed from (3.2.7) to be the kernel of multiplication by t on M. This is a submodule of M, so it is finitely generated if M is, since we are assuming R is left Noetherian, hence R[t] is also left Noetherian by the Hilbert Basis Theorem (Theorem 3.2.1). (ii) We need to show that the relations in Ge are preserved by the map. But this follows immediately from the exact sequence (3.2.8) together with Lemma 3.1.10. (The hypothesis on the category R-Modg needed for the Lemma follows from the assumption that M is left Noetherian.) If R is left regular, then so is R[t] by the Syzygy Theorem (Theorem 3.2.3). Hence by Corollary 3.1.16, the natural maps Ko(R) + Go(R) and Ko(R[tl> -+ Go(W are isomorphisms. The diagram

KdW L

1

Ko(R)

A

Go(R[tl)

1 Go(R)

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3. KO and K1 of Categories, Negative K-Theory

commutes since if M is a finitely generated projective module over R[t], then To$[~](R, M) = 0, hence [M] H [R@R[~I M] under both of the vertical maps in the diagram. Exactly the same reasoning works with R[t] replaced by R[t, t-l], except that now Torr is computed to be the kernel of multiplication by t - 1. The proof for Gr is also almost exactly the same. 0 3.2.10. Corollary. Let R be a left Noetherian ring. Then for j = 0, 1, Gj(R) sits naturally as a direct summand in Gj(R[t]) and in Gj(R[t, t-l]). Proof. If M is a finitely generated R-module, then

To?(R, R[t] @‘R M) and Tar,Wt, ‘- l](R, R[t, t-l] @.R M) * are by (3.2.7) computed as the homology of the complexes R[t] @‘R M 4, R[t] ‘8’R M,

R[t, t-l] @‘R M =+ R[t, t-‘] @‘R M.

So Torr can be seen to vanish and Tore gives back M. Hence the composites Gj CR) + Gj

(WI)

map

of Proposition 3.2.9

b Gj(R)>

are the identity. 0 Now we’re almost ready for the first major result of this section, which is Grothendieck’s Theorem comparing Ge for a ring R and for the ring of polynomials R[t] or R[t, t-l]. It is convenient to begin by first proving the version of the theorem for graded modules. Then we will use a trick to go back to the ungraded case. 3.2.11. Theorem. Let R be a left Noetherian ring, viewed as a graded ring with trivial grading concentrating everything in degree 0. Give the polynomial ring R[tl , . . . , tr] its usual grading in which the elements tl, . . . , t, have degree 1. For a (non-negatively) graded left Noetherian ring, let Grded denote K,-, of the category of finitely generated graded modules M = @nEZ M,. Note that because of the finite generation hypothesis and the fact that the ring is non-negatively graded, these modules are automatically bounded below (i.e., given the module M, there is some ne E Z such that M, = 0 for n < no). Morphisms in this category are required to preserve the grading. Then the exact functor M H R[tl, . . . , tr] @R M induces an isomorphism Go(R) @z Z[t] E Grded(R) -+ Grded(R[tl, . . . , tT]). Proof. First of all, it is obvious that if R is trivially graded, then Go(R) @Q. Z[t] ” Grded(R),

since a finitely generated graded R-module is just a finite direct sum of finitely generated graded modules M,, each concentrated in a single degree, and we identify [M,] E Grded(R) with [M,] @t” in Go(R) @z Z[t].

2. The Grothendieck and Bass-Heller-Swan Theorems

139

Next, observe that everything we have just done works with graded modules as well. In other words, there is a map Grded(R[t]) + Grded(R) defined by sending [M] to [R ~1~1~1 M] - [Torp[tl (R, M)[-1]], where for N a graded module, the symbol N[r] denotes N shifted in degree by T: WI, = K&+7.. The degree shift comes from the fact that in the category of graded modules, the resolution (3.2.7) is not as it stands a resolution by graded modules, since multiplication by t increases degree by 1, but we can replace it by O+R[t][-1] +R[t]+R+O. Corollary 3.2.10 holds in the graded context, and tells us that the map G rded (R[t]) + Grded (R) is a split surjection, with right inverse the map [M] H [R[t] @.R M]. Iterating all of this T times, we see that R[tl, . . . , tr] has finite homological dimension over R, and that there is a split surjection Grded(R[tl,. . . , t,.]) 4 Grded(R) defined using higher Tor’s. For simplicity of notation, let 5’ = R[tl, . . . , t,.]. So it suffices to show that the map Grded(R) --) Grded(S) is surjective. Let 3 be the full subcategory of finitely generated “R-flat” graded S-modules, whose objects are graded modules M satisfying TorT(R, M) = 0 for j > 0. (For instance, when r = 1, these are modules which are ti-torsion-free.) If M lies in this subcategory, the map Graded(S) + GFded(R) takes the simpler form [M] H [R ~3s M]. Because of the long exact Tor sequence, F is a category with exact sequences and contains the kernel of each of its surjective morphisms. So the hypotheses of the Resolution Theorem (Theorem 3.1.13) are satisfied, and the inclusion of F induces an isomorphism &(3) 3 Grded (S). Now if M is a graded R-module, S @R M is R-flat, so the map Gyaded (R) -+ Grded (S) naturally factors through Ks (F), and it’s enough to show that the map GEaded(R) + Kc(F) is surjective. Let M be an object of 3, thus a finitely generated graded S-module, and recall that S is left Noetherian by Theorem 3.2.1. For each integer i, let Fi(M) = the S-submodule of M generated by Mj, j 2 i, Qi = Mi/Mi TI Fi_l(M). Note that Qi is just the component of R CBS M in degree i, since (using multi-index notation) Mi n Fi_l(M) = 2 C tIMi-j. j=l III=j (The sum is really finite, since M is bounded below.) Similarly, R@s Fi(M) vanishes in degrees > i and coincides with R @IS M in degrees 5 i.

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3. Ko and K1 of Categories, Negative K-Theory

If Mi = 0 for i < no, then 0 = Fno_l(M)

c Fno(M) c ... c_ F+cu(M) = M,

and the filtration must terminate at some finite stage, i.e., F,, (M) = A4 for some ni, since M is finitely generated and Noetherian. Note that there is a map of graded S-modules from S @R Mi to F,(M), which induces by passage to the quotient a surjective map of graded S-modules S @‘R Qi + Fi(M)/E-l(M). Here we are viewing Mi and Qi as graded modules concentrated in the single degree i. We will show this map is an isomorphism for each i. For i < no or i > ni, this is obvious since both sides are zero. Suppose we know that Torf(R, Fi(M)) = 0, wh’ic his at least the case for i = n1 since M E Obj FT. From the short exact sequence of graded modules 0 + Fi_l(M) + Fi(M) 4 K(M)/&-l(M) + 0 and the fact that the natural map R @s Fi-I(M) + R 8s K(M) is injective with cokernel Qi, we see first that Torf (R, Fi(M)) = 0 implies also Torf(R, Fi(M)/Fi_l(M)) = 0 and Torf(R, Fi_l(M)) 2 Torf(R, Fi(M)/Fi-l(M)). Then if Ki denotes the kernel of S ‘8R Qi + Fi(M)/Fi-l(M), tensoring with R gives the exact sequence 0 = Torf(R, Fi(M)/Fi-l(M)) + R @S Ki --+ R@s

(S@RQi) +

I/

II

Qi

R 8s (Fi(M)/E-l(M))

=

Qi.

2. The Grothendieck and Bass-Heller-Swan Theorems

141

This shows R 8s Ki = 0, which since Ki is a finitely generated graded module forces Ki = 0. Hence Fi(M)/Fi_l(M) 2 S@RQ~, which shows that Fi(M)/Fi_l(M) E 0bj.F. Substituting back the fact that Torz vanishes, we get Torf(R, Fi_l(M)) = 0. So by descending induction on i, Ki = 0 and Torf (R, Fi(A4)) = 0 for all i, as desired. Now to conclude the argument, note that [M] = Ci[Fi(M)/Fi_l(M)] in Grded(S). But we have seen that Fi(M)/Fi_l(M) g R[t] @.RQ~, hence [Fi(M)/F+l(M)] lies in the image of GrdedR for each i, as required. 0 As we have noted, the next theorem is really due to Grothendieck, though this version of it first appeared in [BassHellerSwan]. 3.2.12. Theorem (Grothendieck). Let R be a left Noetherian ring. Then the natural maps Go(R) + Go(R[t]) and Go(R) + Go(R[t, t-l]) are isomorphisms, with inverses given by the maps of Proposition 3.2.9. Proof. We begin with the csse of R[t]; the case of R[t, t-l] will follow. We need to show the map Go(R) + Go(R[t]) is surjective. The trick is to observe that if M is a finitely generated R[t]-module, then M = R[t] 8.1~ N, where N is a finitely generated graded R[t, s]-module (we give R[t, s] the grading by the total degree of a polynomial) and where + : R[t, s] + R[t] is the surjective homomorphism sending t I+ t, s I+ 1. To see this, note that M = R[t]“/ Q for some module of relations Q G R[tln, and since M is finitely generated and R is left Noetherian, Q is finitely generated because of the Hilbert Basis Theorem (Theorem 3.2.1). Choose a finite set fj =

(fj,l(% . . .,

fj,&>>

7

1 I j I m,

of generators of Q and let d = max deg fj,k. Define

sj = (gj,&

s>, . . . , gj,& 3)) E RIG sln,

1 I j F m,

by replacing each monomial at” in the by atlsdml. Then each gj,k is homogeneous of degree d, and under the map II, : R[t, s] + R[t]. Hence if Q’ is the submodule of R[t, sin generated by the &k’s, N = R[t, s]~/&’ is a finitely generated graded R[t, s]-module and G,(N) = M. Observe in addition that the functor +* from graded R[t, s]-modules to R[t]-modules is exact. Indeed, the tensor product functor is always right exact. On the other hand, 6,(N) may also be written as N/(s - l)N, and we have left exactness because if N is a graded R[t, s]-module and N’ is a graded submodule, and if z = ~~?& xj E N, (s - 1)x E N’, then fj,k’s

gj,k I--+ fj,k

( SX~-1 - xj) E N', j=nfJ+1 that x,,, E N’, x,,+l E N’, . . . , and x E N’. Thus I/J, induces by Proposition 3.1.9 a homomorphism Grded(R[t, s]) + Go(R[t]). SO

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Now consider the diagram

Grded(R) - Grded(R[t, s]) forget grading

**

1

Go(R)

-

1

Go(WlL

where the horizontal arrows are induced by the inclusions of R into R[t] and R[t, s]. This diagram obviously commutes. By Theorem 3.2.11, the top horizontal arrow is an isomorphism. From the fact that every finitely generated R[t]-module is $,(N) for some finitely generated graded R[t, s]module, the right-hand vertical arrow is surjective. Thus the bottom horizontal arrow is surjective and we are done. Now consider the case of Go(R) -+ Go(R[t, t-l]). We must show that this map is also surjective. Since R[t, t-l] is flat over R[t] (as an R[t]module, R[t, t-‘1 = l&n-too t-“R[t], and t-“R[t] is free over R[t]), the inclusion R[t] -+ R[t, t-l] induces a homomorphism Go(R[tl) --+ GoW,

t-l]>

by [M] H [R[t, t-l] BR[~I M]. The map Go(R) + Go(R[t, t-l]) obviously factors through this map. Since we have seen that Go(R) + Go(R[t]) is an isomorphism, we only need to show that Go(R[t]) + Go(R[t, t-l]) is surjective. But if M is a finitely generated R[t, t-‘l-module, M = R[t, t-‘]“/S for some finitely generated module of relations S. (We are using Corollary 3.2.2.) Multiplication by tk induces an automorphism of R[t, t-‘ln, and for large enough k, it will kill off all negative powers of t in a finite set of generators for S. Thus for large enough k, tkS C R[tln, a n d M = tkR[t, t-‘l n/tkS is extended from a finitely generated R[t]module. This shows Go(R[t]) + Go(R[t, t-l]) is surjective, completing the proof. Cl 3.2.13. Corollary. Let R be a left regular ring. Then the natural map Ko(R[t]) -+ Ko(R) induced by (3.2.5) and the natural map Ko(R[t, t-l]) + Ko(R) induced by (3.2.6) are isomorphisms. Alternatively Ko(tR[t]) and Ko((t - l)R[t, t-‘I) C cornP u te din the sense of Ko for rings without unit)

vanish.

Proof. This follows from combining Theorem 3.2.12, Corollary 3.1.16, and Proposition 3.2.9. q Remark. For rings which are not left regular, the maps Ko(R[t]) + Ko(R) and Ko(R[t, t-l]) + Ko(R) can have a non-zero kernel. For an

example of the former phenomenon, see Exercise 3.2.24. The kernel of the map &(R[t, t-l]) 4 Ko(R) actually consists of two different parts, both of which can be non-zero. The first is related to the kernel of Ko(R[t]) + Ko(R); the second is the functor K-l(R) which will be studied in the next section.

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3.2.14. Definition. If R is any ring with unit, we define NKj(R), j = 0 or 1, to be the relative K-group Kj(R[t], tR[t]). By the split short exact sequence (3.2.5), this is the same as the kernel of the map on K-theory induced by R[t] 2 R. (Recall Exercises 1.5.11 and 2.5.19.) Corollary 3.2.13 states that NKo(R) vanishes if R is left regular. Next we come to the study of K1 and G1. We would like to show as in the case of Kc that Gl(R[t]) Z Gl(R) for R left Noetherian, so that NKl(R) vanishes if R is left regular. The case of Laurent polynomials will now be a bit different, since as we remarked at the beginning of this section, there is reason to believe Kl(R[t, t-l]) should be related to Kl(R) @ Ko(R), not just to Kl(R). First we have the analogue of Theorems 3.2.11 and 3.2.12 for GI. 3.2.X. Theorem. Let R be a left Noetherian ring, viewed as a graded ring with trivial grading, and Jet R[tl, . . . , t,.] be given its usual grading. For a (non-negatively) graded left Noetherian ring, Jet Gyded denote Kl of the category of finitely generated graded modules M = BnEZ M,, . (Morphisms in this category are required to preserve the grading.) Then the exact functor M H R[tl, . . . , tT] ‘8~ M induces an isomorphism Gl(R) @Z Z[t] Z Gyded(R) -+ Gytied(R[tl,. . . ,&I). Proof. For simplicity of notation, let S = R[tl, . . . , t,.]. As in the proof of Theorem 3.2.11, it suffices to show that the map Gyded(R) -+ GFaded(S) is surjective. As in the proof of Theorem 3.2.11, let 7 be the full subcategory of the category of finitely generated R-flat graded S-modules, whose objects are graded modules M satisfying Torf(R, M) = 0 for j > 0. These include the finitely generated free graded modules, and any graded morphism lifts to a morphism of a free graded module. As in the proof of Theorem 3.2.11, the hypotheses of the Resolution Theorem (Theorem 3.1.14) are satisfied, and the inclusion of T induces an isomorphism K1 (F) -% Gyded(S). Also the map GFded(R) --) Gyded(S) naturally factors through K1(9), and it’s enough to show that the map Grded(R) + Kl(3) is surjective. Furthermore, by the method of the proof of Theorem 3.2.11, it is enough to consider classes in Kl(3) defined by an automorphism (Y of S ‘8~ M, where M is a graded R-module. Then since cr is required to be gradingpreserving, (Y induces an automorphism of (S @R M)n = M, and since M generates S@R M as an S-module, Q is determined by its restriction to M, i.e., (Y = l@‘cy[~. Thus the map Gyded(R) -+ Gyaded(S) is surjective. El 3.2.16. Theorem (Grothendieck Theorem for GI). Let R be a left Noetherian ring. Then the natural map Gl(R) --) Gl(R[t]) is an isomorphism, with inverse given by the map of Proposition 3.2.9.

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3. Ko and K1 of Categories, Negative K-Theory

Proof. As in the proof of Theorem 3.2.12, we consider the diagram

Grded(R) - Gyded(R[t, s]) forget grading

6

1

G(R)

-

1

G(RM),

where the horizontal arrows are induced by the inclusions of R into R(t] and R[t, s]. This diagram obviously commutes. By Theorem 3.2.15, the top horizontal arrow is an isomorphism. So it will be enough to show that the vertical arrow on the right is surjective. This is a bit more delicate than in the case of Ge since we need to consider not just modules but also their automorphisms. But ?I, : R[t, s] + R[t] factors through the inclusion R[t, s] L) R[t, s, s-l], so closer examination shows that the right-hand side of the diagram above can be factored as

Gyded(R[t, s]) A4 11.

yded(R[t, s, s-l])

1 JG

G1(R[tl), where Gyded(R[t, s, s-l ]) is defined using Z-graded modules that are not necessarily bounded below. (If N is a graded R[t, s, s-‘l-module, then multiplication by s induces isomorphisms Nj + Nj-1 for all j, so N can’t be bounded below unless it’s the zero module.) Furthermore, if M is an R[t]-module, then R[s, s-l] @IR A4 can be given the structure of a graded R[t, s, s-‘]-module F(M) ( in which t acts by the original action of t composed with multiplication by s), and the functors $J* and F are inverses to one another, defining an equivalence of the category of finitely generated R[t]-modules with the category of finitely generated graded R[t, s, s-l]modules. So G yded(R[t, s, SC’]) + Gl(R[t]) is an isomorphism and we need only see that G yded(R[t, s]) + Gyded(R[t, s, s-l]) is surjective. Thus let N be a finitely generated graded R[t, s, s-‘]-module and let cr be a grading-preserving automorphism of N. Let P = @r=, iV, be the R[t, s]-module generated by No. Then P is finitely generated and (Y maps P into itself, and N = R[t, s, s-l] @R[t, so P. Since obviously (Y = 1 @((YIP), this shows [N, o] is in the image of Gyded(R[t, s]). So G yded(R[t, s]) + Gyded(R[t, s, s-l]) is surjective. 0

2. The Grothendieck and Bass-Heller-Swan Theorems

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3.2.17. Corollary. If R is a left regular ring, then the natural map Kl(R[t]) + Kl(R) is an isomorphism. In other words, NK1 (R) = 0. Proof. This follows from combining Theorem 3.2.16, Corollary 3.1.16, and Proposition 3.2.9. q We come now to the case of the Laurent polynomial ring R[t, t-l]. We begin with Gi and then go on to KI. 3.2.18. Proposition. Let R be a left Noetherian ring. Then there is a natural embedding of G1 (R) @ Go(R) as a direct summand in Gl(R[t, t-l]) via + : ([M,

a], [M’])

- [R[t, t-l] ‘8’~ M, 163

a] + [R[t, t-l] ‘8’R M’, t C3 I] .

The left inverse Q to @ is given by Gl(R[t, t-l]) -+ Gl(R) as defined in Proposition 3.2.9, together with the following map Gl(R[t, t-l]) -+ Go(R): Let N be a finitely generated R[t, t-l]-module, p E Aut N, and Jet N’ be a finitely generated R[t]-submodule of N that generates N as an R[t, t-I]-module. Then for suitably large Jc, t”P maps N’ into itself and coker((t’p)]N/) is a finitely generated R-module. Map [N, /3] E Gl(R[t, t-l]) to [coker((t”P)]Nr)] - k[coker(tJNt)] E Go(R). Proof. It is clear that Cp defines a homomorphism, and we already verified in Corollary 3.2.10 that it embeds Gl(R) in Gl(R[t, t-l]) as a direct summand. It therefore suffices to check that the indicated formula gives a well-defined homomorphism Gl(R[t, t-l]) + Go(R), and that the composite Q o @ is the identity. The first problem is to show that, given a finitely generated R[t, t-l]module N and ,f3 E Aut N, [coker((t”fl)]Nt)] - k[coker(t]nrt)] E Go(R) is independent of the choice of N’ and of k. First of all, suppose N’ is fixed and t”/3 maps N’ into itself. Note that (tk/31Nt) is an injective R[t]-module homomorphism since it is the restriction of an automorphism of N. If we replace k by k+j, j > 0, then t”+j/?(N’) C t”P (N’) c N’, and tk/? induces an iaomorphism N’/ tjN’ -+ t”P (N’) /t”+ j/3(N’) . Hence, in Ge( R), we have [coker((tkp)]Nr)] + [coker(tj]Nr)] = [coker((t”+jp)]Nf)]. In particular, iterating this with /3 = 1 shows that [coker(tj]Nt)] = j[coker(t]Nl)], and so [coker((t”+j/?)]N~)]--(k+j)[coker(tJNr)]

= [coker((tkp)JNt)]-k[coker(tJNj)].

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3. KO and K1 of Categories, Negative K-Theory

Thus for fixed N’ we have independence of k. Now suppose N” is another finitely generated R[t]-submodule of N that generates N as an R[t, t-l]-module. Then N=

z

t-jN’ =

j=o

5 t+N”, j=o

so for suitably large j, tj N” C N’. If we choose k large enough so that t”,f? maps N’ and N” into themselves, then (tj+lep)N” C tjN” n (t”, L?)N’ 2 N’ and in Gc( R) we have [N’/ (tk@N’] + [(t”/ 3)N’/ (tj+“@ N”) ] = [N’/ tjN”] + [tjN”/ (tj+k/3)N”]

or [N’/ (t’P )N’] + [N’/ tjN”] = [N’/ tjN”] + [N”/ (tk@)N”] , so [N’/ (tk,d)N’] = [N”/ (t”p )N”] . This proves independence of the choice

of N’ and shows 9 is well defined. Next we have to show that \E is a homomorphism. If fi and y are two automorphisms of N, and if we choose k large enough so that t”P and t”y both map N’ into itself, then (t2”@y)N’ 2 (tkp)N’ fl (tky)N’ c N’, so in Go(R) we have

[N’/(t2kPr)N’]

= [N’/(t”y)N’]

+ [(tkr)N’/(t2”/3r)N’]

= [N’/(t”y)N’]

+ [N’/(tk@N’],

showing that Q([N, ,&]) = Q([N, /?I) + 9([N, 71). It remains to show that 9 is additive on short exact sequences. Suppose

is a short exact sequence of finitely generated R[t, t-l]-modules, and p is an automorphism of N2 that maps Ni onto itself and that induces an automorphism y of Ns. Choose a finitely generated R[t]-submodule Ni of N2 that generates N2 as an R[t, t-‘l-module. Let N: = Ni n Ni, and let Ni be the image of Nl in Na. Then Ni and Ni are finitely generated and generate Ni and Ns, respectively. If we choose k so that tkp maps Ni and Ni into themselves, then if z E Ni n (t”P )Ni, we have (t”P)-‘cc E NI n Nl (since Ni is stable under (t”P)-‘), which is Ni. Hence Ni n (tk,B)Ni = (t”P )Ni and so in Go(R),

P%l(tkW$l = Pi/ (N; + (tkPP%)l + [(N; + V”PP;) &“P>N;11 = P%l(t”4Nil+ Pi/ (N: n @“PP’$)l = [N;/(tkr)N;l + [N:I(t”P)N;l. It follows that Q([Ns, 71) + XP([Ni, PIN,]) = *([Nz, P]), so 9 is additive on short exact sequences.

2. The Grothendieck and Bass-Heller-Swan Theorems

147

Finally, we show that the composite Q o @ is the identity. If M and M’ are finitely generated R-modules and cx E Aut M, then \k o Cp ([M,

a], [M’]) = ‘32 ([R[t, t-l] @R M, 163 a] + [R[t,

t-l]

@‘R M', t @ l])

= ([M, 4 + [M’, 11, [coker(l@ ~)IR[~J@~M] + [coker(t@ ~)(R[~J~~M~]) = ([M, aI +O, 0 + [M'l) = ([M, a], [M']),

as required. 0 3.2.19. Theorem. Let R be a left Noetherian ring. Then the embedding Q ofGl(R)@Go(R) into Gl(R[t, t-l]), defined in Proposition 3.2.18, is an isomorphism. Proof. We need to show that @ is surjective. The intuitive idea is easy to explain. Suppose [N, p] E Gl(R[t, t-l]) is defined by a finitely generated R[t, t-l]-module and p E Aut N. We need to show that if [N, PI +-+ 0 in Go(R), then [N, PI comes from a class in Gr (R). Let N’ be a finitely generated R[t]-submodule of N that generates N as an R[t, t-‘l-module, and suppose for simplicity that p maps N’ into itself. The statement that [N, fl] H 0 in Go(R) then means that [N’/P(N’)] = 0 in Go(R). If N’/P(N’) were literally the zero-module, this would mean that p restricts to an automorphism of N’. But then N = R[t, t-l] @R[t] N’ and P = 1 C%J P(w, which shows that [N, p] lies in the image of the map Gi(R[t]) + Gl(R[t, t-l]). S ince (using Theorem 3.2.16) we have a commutative diagram Gl (R)

G WI) - G(R[t, t-l]>, this shows [N, p] is in the image of Gl(R). To make this argument rigorous takes a bit of work, and can be done in a number of ways. The easiest is probably to appeal to the method of proof of Theorem 3.2.16, which shows that we can take N’ = R[t] ~3.1~ P, with P a finitely generated graded R[t, s, .sml ]-module, and that /3 extends to a graded automorphism of R[t, t-l, s, s-l] @R[t, S, S-~~ P. Instead of assuming that [N, p] H 0 in Go(R), we’ll make no assumption on /3 and show how to write [N, p] in terms of elements in the image of a. First multiply p by a suitably high power of t so that p maps N’ into itself. Then there will be an induced grading-preserving endomorphism fi of P. Let P’ = @,“= , P, be the graded R[t, s]-submodule of P generated by the elements of degree 0; a maps P’ into itself. By the method of proof of Theorem 3.2.15, we may reduce to the case where P’ = R[t, s] @R M, M a finitely generated graded R-module, and /? is determined by a graded endomorphism a of M. But then N = R[t, t-l] @R

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3. Ko and K1 of Categories, Negative K-Theory

M and /3 induces (Y on the quotient M Z N/(t - l)N. Since /3 is invertible, cx is in fact an automorphism of M. We may identify N with enEZ t”M and write @(tncc) = xjEz&(z)t n+i for z E M, where almost all of the pi(x) are 0, and Cj &(x) = CX(X). From this one can see M is an iterated

extension of R-modules Mi on which /3 takes the simple form tjcxj, and we have [N, ,L?] =

C [R[t, t-‘] ‘8 R Mi, t’ @ aj]

jCZ = C

([R[t, t-l] 8~ Mj, ti @ l] + [R[t, t-‘1

‘8 R Mj, 18 aj])

jEZ

which shows @ is surjective.

Cl

3.2.20. Corollary (Bass-Heller-Swan). Let R be a left regular ring. Then there is a canonical isomorphism K1 (R[t, t-l]) E ICI(R) @ KC,(R). Proof. This follows from combining Theorem 3.2.19, Corollary 3.1.16, and Proposition 3.2.9. Cl Since we are also interested in rings which are not left regular or even left Noetherian, it will be convenient to try to analyze Kl(R[t, t-l]) directly, without going through the intermediary of G-theory. This will lead to another proof of the Bass-Heller-Swan Theorem (Corollary 3.2.20) as well as a motivation for the definition of negative K-theory in the next section. 3.2.21. Lemma ([BassHellerSwan]). Let R be a ring. Then (4 Any matrix B E GL(R[t]) can be reduced, module GL(R) and E(R[t]), to a matrix of the form 1 + At, where A is a nilpotent matrix with entries in R. (b) Any matrix B E GL(R[t, t-l]) can be reduced, module GL(R) “i y (1 + A(t - l)), and E(R[t, t-l]), to a matrix of the form ( ) where A is a matrix with entries in R and A(1 - A) is nilpotent. Proof. (a) Write B = Bo + tB1 + . . . + tdBd, where the Bj are matrices with entries in R. We will first show by induction that B can be reduced to something with d 5 1. So assume d > 1. Then if N stands for “is equal to, modulo GL(R) and E(R[t]),” we have

2. The Grothendieck and Bass-Heller-Swan Theorems

149

(since we can add a multiple of the bottom row to the top row)

(since we can subtract tx (the last column) from the first column). Now we have something of degree 5 d - 1. Continuing by induction, Bean be reduced to something with d < 1. If we can reduce to d = 0, the assertion of the Lemma is obvious. Otherwise, we can reduce to the case d = 1, and assume B = Be + tB1. Since B is invertible as a matrix over R[t], Be must be invertible. Factoring out Be, we reduce B to the form 1 + At. This must be invertible as a matrix overR[t],sowehaveB-l=Cs+tCr+... + t’C, for some Cj ‘S and some integer r. Multiplying out the equation 1 = (1 +At)(Cc +tCl +...+t’C ,) = (Co+ tC1 +...+t’C ,)(l +At), we obtain the equations

1 = CrJ,

0 = ACc + Cr = CeA + Cl,

0 = AC,_1 + C, = C&IA + C,.,

...,

0 = AC,. = C,A.

Solving inductively, we obtain Cc = 1, Cr = -A, . . . , Cj = (-A)j, and A is nilpotent since AT+l = 0. (b) Since we are allowed to multiply B by a power oft, we may suppose B involves only non-negative powers of t. Then we may repeat the same trick and come down to the case where B = Be + tB1 = (Be + Br) + (t - 1)Br. Since B is invertible as a matrix over R[t, t-l], Be + B1 must be invertible. Factoring out Be + B1, we reduce B to the form 1 + A(t - 1) = 1 - A + At. This must be invertible as a matrix over R[t, t-l], so after multiplying by a power of t it has an inverse which is a matrix over R[t]. By the same reasoning as in (a), (1 - A)A is nilpotent. 0 3.2.22. Theorem (Bass-Heller-Swan). Let R be a ring. Let Nil R be the category whose objects are pairs (P, A), where P is a finitely generated free R-module and A is a nilpotent endomorphism of P. The morphisms (P, A) + (P’, A’) are R-module homomorphisms T : P + P’ such that A’T = TP. Note that Nil R is a category with exact sequences, and Ke of this category contains an obvious homomorphic image of Z coming from the full subcategory of objects with A = 0. Then

(4 (b)

Kr (R[t]) = Kr (R) $ NK1 (R), where NK1 (R) is canonically iso morphic to &(Nil R). (The notation Kc means Ke divided out by the canonical image of Z.) There is a natural splitting of Kl(R[t, t-l]) as Kl(R) @Kc(R) @ NKl(R) $ NK1(R). The two copies of NKr(R) come from the embeddings R[t] c-) R[t, t-‘1 and R[t-‘1 L) R[t, t-l].

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3. KO and K1 of Categories, Negative K-Theory

Proof. (a) We already know there must be a splitting

and (a) of Lemma 3.2.21 shows that NKr(R) is the image in Kr(R[t]) of matrices of the form 1 + At, A nilpotent. We define a map NKr(R) + I?e(Nil R) by (when A is a nilpotent n x n matrix) NK1(R) 3 [l + At] H [(R”, A)] E &,(Nil R). To check that this is well defined, note that if 1 + At is conjugate to 1 + A’t under GL(n, R[t]), then sending t I-+ 1, we see 1 + A is conjugate to 1 + A’ under GL(n, R), hence A is conjugate to A’ under GL(n, R) and [(R”, A)] = [(R”, A’)] in Ke(Nil R). Furthermore, if we replace l,+At by the (n+Ic) x (n+k) matrix (ln+At)@(lk), this corresponds to replacing A by A@Ok and [(R”, A)] by [(R”, A)] + [(R’“, O)]. This element is different in Ko, but the same in &, so the map NKl(R) -+ &(Nil R) is well defined. Finally, the map is a homomorphism since [l + At] + [l + A’t ] = [(l + At) @ (1 + A’t )] = [l + (A @ A’)t]

++ [(R”, A)1 + [(Rn,

-4’11.

To show the map is an isomorphism, we construct its inverse by the obvious formula [(R”, A)] H [l + At]. Note that [l + At] indeed defines a class in NKl(R), since for any nilpotent A, 1 + At E GL(R[t]) and maps to 1 in GL(R) under the homomorphism defined by t H 0. Since we know all classes in NKr (R) can be represented in the form [l + At], we will be done if we can show that the map is indeed well defined. Clearly the class of 1 + At doesn’t change if we replace A by A @ 01, (which corresponds to changing our class in &(Nil R) by something in the canonical image of Z). So we only need to check additivity on exact sequences. Suppose

is a short exact sequence of finitely generated free R-modules and we have nilpotent endomorphisms Aj E End(Pj) such that AP(YI = alAl, Asa~ = azA2. This means that also (1 + A2 t) a1 = a~(l + Alt) and (1 + Ast)az = as(1 +

A&, so [l + A& = [l + Ad] + [l + Ad]

in Kl(R[t]),

and we’re done. (b) The maps Cp and \E from Proposition 3.2.18 can be defined in Ktheory instead of in G-theory, by exactly the same formulas. The only point that needs checking is that the second component of * indeed sends Kl(R[t, t-l]) into Ko(R) and not just into Go(R). To check this, we use the fact that the cokernel of the map Kl(R) -+ Kl(R[t, t-l]) is described by

2. The Grothendieck and Bass-Heller-Swan Theorems

151

Lemma 3.2.21(b) as being generated by matrices of the form l+A(t-1) with A an n. x n matrix over R and (1 - A)A nilpotent. Equivalently, A = P+N, where P is idempotent, N is nilpotent, and P and N commute with each other. To see this, suppose A’( 1 - A)’ = 0. Then since the polynomials zT and (1 - z)’ are relatively prime in Z[z], there are polynomials p(z) and q(z) with integer coefficients such that p(z)zY + q(z)(l - XC)’ = 1. Let P = p(A)A’. Then 1 - P = q(A)(l - A)’ and since A’(1 - A)’ = 0, P(l - P) = 0. This shows P is idempotent, and P is a polynomial in A. If N = A - P, then N is also a polynomial in A, so P and N commute with one another. Furthermore, N = A (1 - p(A)A’-l) and N = (A - 1) + (1 - P) = (1 - A) (-1 + q(A)(l - A)T-l), so N is divisible by both A and 1 - A, hence divisible by A(1 - A), hence nilpotent. But then R[t]“/ (1 + A(t - 1)) R[tln E im P is projective as an R-module. So Kr (R) @ Ko(R) naturally embeds as a direct summand in Kr (R[t]). The cokernel of this embedding is once again described by Lemma 3.2.21 (b) as being generated by matrices of the form 1 -I- (P + N) (t - l), where P is idempotent, N is nilpotent, and P and N commute with one another. Thus 1 + (P + N)(t - 1) corresponds to a pair of nilpotent matrices, PN and (1 - P)N. These correspond to the two copies of NKr (R). The rest of the proof is just as in part (a). Cl Remark. This explains a commonly used notation: the group NKl(R) is often called Nil R because of (a) of the theorem. 3.2.23. Exercise (Non-triviality of NK1). Let k be a commutative field, and let R = k[t]/(t2). (a) Show that R is a local ring and thus compute Ko(R) and Kl(R). (b) Let s be another indeterminate and compute the group of units R[s] x in R[s]. (c) From the exact sequence (split on the right) NKl(R) -+ KI(R[s]) s Kl(R), deduce that NK1 (R) is not finitely generated (as an abelian group). (Recall that since the ring R[s] is commutative, R[slX -+ Kl(R[s]).) 3.2.24. Exercise (Non-triviality of NKo). Let k be a commutative field, and let S = k[t2, t3]. (a) From the split exact sequence 0 + t2k[t] -+ S % k + 0

and the long exact K-theory sequence coming from the short exact sequence 0 + t2k[t] + k[t] + k[t]/(t2) --f 0,

compute Ko(S). (Hint: use Exercise 3.2.23(a) and the fact that k[t] is a Euclidean ring.)

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3. KO and K1 of Categories, Negative K-Theory

(b) Let s be another indeterminate and similarly use the short exact sequences 0 -+ t%[t,s] + S[s] % Ic[s] + 0 and 0 -+ t%[t, s] --+ /+, s] + Ic[t, s]/(t2) 4 0 to relate Kc(S[s]) and thus NKs(S) to NKi(R) in Exercise 3.2.24(c). (Recall that Ke (S[s]) g Ks (S)@NKc (S) .) Use Grothendieck’s Theorem which implies that Ke(lc[t, s]) = 0. Deduce that NKc(S) is not finitely generated (as an abelian group). 3.2.25. Exercise. Give another proof of Corollaries 3.2.17 and 3.2.20 from Theorem 3.2.22, by showing that the group NK1 described in that Theorem has to vanish if R is left regular. (Hint: if A is an n x n matrix over R and A’ = 0, it gives a filtration of the free module R” by ‘C_... O=imArCimAr-

&imAc Rn

If R is left regular, the subquotients can be resolved by finitely generated projective modules.) 3.2.26. Exercise. Let R be any ring. (1) Show that NKl(R[t, t-l]) contains NKl(R)@NKo(R) as adirect summand. You can do this by computing Kl(R[t, t-l, s]) two ways. (2) [Vorst] A ring R is called j-regular if Kj (R[tl, . . . , tT]) E Kj (R) for any T. By Corollary 3.2.17, a left regular ring is l-regular. Show that NKl(R[t]) = 0 implies NKl(R[t, t-l]) = 0, by noting that a nilpotent matrix over R[t, t-l] is of the form tpn x (a nilpotent matrix over R[t]). Then deduce from (1) that if R is l-regular, NKo(R) = 0. (3) Iterating the result of (2), prove the multivariable version of this, that l-regularity implies O-regularity. (However, there are O-regular rings which are not l-regular.) 3.2.27. Exercise. (1) Show from the Bass-Heller-Swan Theorem that if 7r is any group, then Wh(rxZ) Z Wh(7r)@Ke(Z7r)$(NKi(Zrr))2. (We are mixing additive and multiplicative notation here.) (2) Deduce that the Whitehead group of any free abelian group vanishes. (Hint: Z[tl, tl’, . . . , t,, t;‘] is regular. Why?) (3) See if you can find a topological interpretation of the formula in (l), and in particular a relationship between Wall finiteness obstructions for spaces with fundamental group r and Whitehead torsion obstructions for spaces with fundamental group 7r x Z. 3.2.28. Exercise. Let k be a (commutative) field. Show that Corollary 3.2.20 applied with R = k amounts to the assertion that SK1 vanishes for the PID k[t, t-l]. Can you prove this directly using the results of Chapter 2?

3. Negative K-theory 153

3.2.29. Exercise [Farrell]. Let R be a ring, and view NKl(R) as being represented by classes of nilpotent matrices N over R, as in Theorem 3.2.22. (1) Fix an integer n, and let L, : R[P] L) R[t] be the inclusion. Note that R[t] is a free R[t”]-module of rank n, and thus that an T x T matrix over R[t] gives rise to an rn x rn matrix over R[P]. In this way a transfer map L: : Kl(R[t]) -+ Kl(R[t”] ) is defined. Show that L: o (Lo)* is multiplication by n on KI(R[P]) (if we use additive notation). (2) Suppose N is a nilpotent r x r matrix over R, so that 1 + Nt represents a typical element of NKr (R). Show that L:( 1 + Nt) is given by the block matrix

0 0 .-. 1 0 0 0 . . . N 1 (3) Let A be the strictly lower-triangular block matrix

A =

1 0 ... Nl 0 ’. ON .

0

0 0 ;

0 0

1 N

0 1

0 0

.*. ...

1.

Show that if N” = 0, then A-‘M is strictly upper-triangular and hence elementary, and thus that ~t([l + Nt]) = 0 in Kl(R[t”] ). Conclude that if NKl(R) is finitely generated, then there is some integer no such that L: = 0 on NKi (R) for all n 2 no. (4) Using (l), deduce that if there is a prime p such that multiplication bypisinjectiveon NKl(R), thenLzo(L,),([l+NP]) # Oforn =pi a power of p and for all nilpotent matrices N. (5) Conclude from (3) and (4) that if NKl(R) is finitely generated and non-zero, there can be no prime p such that multiplication by p is injective on NKl(R), which contradicts the structure theorem for finitely generated abelian groups. Therefore if NKl(R) # 0, then NKi(R) is not finitely generated.

3. Negative K-theory One immediate consequence of the Bass-Heller-Swan Theorem (Theorem 3.2.22) is that for any ring R, Ko(R) = coker (Kl(R[t]) @ K1(R[tel]) --+ Kl(R[t, t-l])) .

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3. KO and K1 of Categories, Negative K-Theory

This motivates the following inductive definition: 3.3.1. Definition. For any ring R, K-l(R) is defined to be the cokernel of the natural map Ko(R[t]) @Ko(R[t-l]) -+ Ko(R[t, t-l]). (Since we have defined Kc even for rings without unit, it is not necessary here to assume that R has a unit.) By Corollary 3.2.20, K-l(R) vanishes if R is left regular. Then for any n > 1, we define K_,(R) to be the cokernel of the natural map K_(,_1)(R[t])@K_(,_1)(R[t-‘]) + K_(,_l)(R[t, t-l]). Note that this is functorial in R, since K_,(R) is a natural direct summand in Ko(R[ti, t;‘, . . . , t,, t,‘]). We a1 sod efine NK_,(R) to be the cokernel of the natural map K_,(R) + K_,(R[t]). (Because of the splitting map R[t] --t R sending t I+ 0, K_,(R[t]) splits as K_,(R) @ NK_,(R).) By iterated use of the Syzygy Theorem (Theorem 3.2.3) and Corollary 3.2.20, K_,(R) and NK_,(R) vanish for all n if R is left regular. The following theorem shows that Theorem 3.2.22 has an exact analogue for Ko, using the new functor K-1. 3.3.2. Theorem. For any ring R, there is a natural splitting Ko(R[t, t-l]) = Ko(R) CB K-I(R) CD NKo(R) @ NKo(R), where the two copies of NKo(R) come from the embeddings R[t] w R[t, t-‘1 and R[t-‘1 L) R[t, t-l]. Proof. Theorem 3.2.22 says that for any ring, Ko(R) naturally sits as a direct summand in K1 (R[tl, t,‘]). It also says that for any ring S, there is a natural exact sequence

0 + Kl(S) + Kl(S[tl) @

K1(W1l) + Kl(S[t, t-l]),

and the cokernel of the map on the right splits. Let us put these two statements together, but taking S = R[tl, t,‘]. Then Ko(R) naturally sits as a direct summand in K1 (S), and similarly Ko(R[t]), Ko(R[t-‘I), and Ko(R[t, t-l]) naturally sit as direct summands in Kl(S[t]), in Kl(S[t-l]), and in Kl(S[t, t-l]), respectively. Furthermore, the diagram 0 ---+ K,,(R) - Ko(R[t]) $ Ko(R[rl]) -

0 - KI(S) - Kl(S[tl) @K1(S[t-‘I) - Ko(R[t,

t-l])

+ K-l(R) -

+ K1(S[t, t-l]) =

0

Ko(S) - 0

3. Negative K-theory 155

clearly commutes. The bottom row is exact, and the top row is exact on the right by definition of K-r(R) and on the left since Kc(R) --+ &(R[t]) is split injective. Let us show that the top row is also exact at Ke(R[t]) @ KO(R[t-l]), and that the top exact sequence splits on the right. To prove the first of these statements, note that by commutativity of the diagram, the kernel of Ko(R[t]) CB Ko(R[t-l]) -+ Ko(R[t, t-l]) may be identified with the intersection in Kr(S[t]) CB Kr(S[t-l]) of the images of Ko(R[t]) @ KO(R[t-l]) and of Kr(S). This is obviously Ko(R). To construct a splitting m a p K-r(R) + Ko(R[t, t-‘l), note that the projection of Kr(S[t, t-l]) onto Kr (S) @ NKr (S) @ NKr (S) killing Kc(S) restricts to a projection of Ko(R[t, t-l]) onto Ko(R) CD NKo(R) 6~ NKo(R) killing K-l(R), hence the latter must split off as a direct summand. 0 Note that iteration of the same argument clearly proves the following. Theorem (“F’u ndamental Theorem of Algebraic K3.3.3. Theory”) . For any ring R and any n 2 1, there is a natural splitting K-(,-r)(R[t, t-‘I) 2 K-(,-r)(R) @K-,(R) @ NK-(,-r)(R) @ NK-(,-r)(R), where the two copies of NK_(,_l)(R) come from the embeddings R[t] -+ R[t, t-l] and R[t-‘1 L) R[t, t-l].

The advantage of the construction of the functors K_,(R) is that it now gives us a way of extending the exact sequence of an ideal arbitrarily far to the right, and thus a way of computing Ko(R/I) from information about R and the ideal I. 3.3.4. Theorem. Let R be a ring with unit, and let I be a two-sided ideal in R, viewed as a ring without unit. Then the exact sequence of Theorem 2.5.4 extends to an exact sequence ... --) Ko(R) 3 K,,(R/I) 2 K_1(I) 4; K_1(R) % K_l(R/I) -% K_z(I) k ... , where L* and q* are the maps induced by the inclusion L : I - R and by the quotient map q : R + R/I. Proof. Take S = R[t, t-‘1 and J = I[t, t-l]. To avoid unnecessary extra notation, again denote the inclusion J q S by L and the quotient map S + S/J Z (R/I)[t, t-l] by q. Then J i S, and by Theorem 2.5.4, there is a natural exact sequence

Kr(S) 3 Kl(SIJ) 5 KO(J) 4; Ko(S) % KO(S/J). On the other hand, we have natural embeddings of Ko(R) in Kr (S), Ko(R/I) in Kl(S/J), K-I(I) in Ko(J), etc., as direct summands, and Kr(S) -% KI(SIJ)

Ko(R)

a Ko(RII),

K , , ( J ) 4; K , , ( S ) 3 Ko(S/J)

K_1(I) 4; K-l(R) 3 K-l(R/I)

156

3. KO and K1 of Categories, Negative K-Theory

commute. columns 0

We also have a commutative diagram with exact rows and - Ki(R) -

Kl(W @ Kl(W1l)

4.

Q* 1

1

II

0 - Kl(RII) - Kl((RIl)[tl)~K1((RIl)[t-lI) a 0

-

1

&(I)

-

Ko(W’l)

L* 1 1 0 - Ko(R) - Ko(W) @ ~o(W’l> 9. 1 Q* 1 II 0 - Ko(R/I) - Ko((WQPl) @ ~o((w~)~~-‘l) II

-

a 1

II

-

Ko(W]) @

-

‘*

...-

Kl(W,

t-l])

= &(R) -

0

1 I . . . - K1((R/I)[t, t-l]) = &(R/I) - 0 a1 . . . - Ko(W, t-l]> = K-1(1) - 0I L* 1 I/ * * * - Ko(R[t, t-l]) = K-1(R) - 0 Q. 1 I * . . - Ko((R/l)[t, t-l]) = K_1(R/I) - 0. Q.

A diagram chase now gives the desired exact sequence as far as the K-I(WI), butwe can iterate the construction to include K-2 terms and eventually K_, terms for all 0. Cl 3.3.5. Examples. (Cf. Examples 1.5.10 and 2.5.6.) (a) Suppose R = Z and I = (m), where m > 0. Then R/I is a product of k local rings, where k is the number of distinct prime factors of m, and we determined earlier that Ko(R/I) Z Zk and that the map Ko(R) -+ Ko(R/I) may be identified with the diagonal map Z + Z”. The cokernel of this map is free abelian of rank k - 1. Since R is a PID, it is certainly a regular ring, hence K-i(R) = 0. Therefore the exact sequence of the ideal terminates with Kc(R) + Ko(R/I) + K-l(I) --+ 0, and K_ i (I) is free abelian of rank k - 1. Looking at the rest of the exact sequence (and using the fact that all negative K-groups of R must vanish) also shows that

3. Negative K-theory 157

K_,(R/I) E K-,-l(I) for n 2 1. We can use this fact to compute the rest of the negative K-groups of I. For instance, suppose m is square-free, i.e., a product of distinct primes. Then R/I is a product of fields, hence is left regular, so all the negative K-groups of R/I must also vanish. Hence K_,(I) = 0 for n > 2, though K-i (I) will be non-zero if Ic > 1. If, say, m = p2 with p prime, then R/I is local but not regular (see Exercise 3.1.24). Thus K-i(I) = 0 from the exact sequence, but, at least a priori, R/I could have plenty of negative K-groups. However we have an exact sequence 0 -+ m -+ Z/(p”) + F, --+ 0, where m = pZ/(p”) is the maximal ideal of Z/(p2). Since m2 = 0, it is easy to see that for any n, m[ti, t;l, . . . . t,, t;‘] = radZ/(p2)[ti, t;‘, . . . , t,, t;‘], with quotient lFP[ti, t;‘, . . . , t,, t;‘] . By the argument of Theorem 1.3.11 (see Exercise 3.3.6 below), the map Ko (zl(~~)[ti, tT1, . . . , Gz, t,‘]) +Ko (@‘&, tl’, . . . , tn, t,‘]) rz (by iterated use of Corollary 3.2.13) is injective, hence all negative K-groups of R/I vanish. Hence K_,(I) = 0 for all n 2 1. (b) Suppose G is a cyclic group of prime order p, say with generator t, and R = ZG is its integral group ring, which may be identified with Z[t]/(tp - 1). If 5 = e2KilP, a primitive pth root of unity, and if S = Z[[], then S is the ring of integers in the cyclotomic field Q(E), hence is a Dedekind domain by Theorem 1.4.18. There is a surjective homomorphism R ++ S defined by sending t - 1. From the exact sequence Ko(R) -+ Ko(S) + K-l(I) = 0, we conclude that the map Ko(R) -+ Ko(S) is surjective. On the other hand, by Corollary 2.5.9, the ma? Ko(R) --+ Ko(S) is injective, so we conclude that Ko(ZG) E Ko(Z[ 0.

158

3. KO and K1 of Categories, Negative K-Theory

It is perhaps worth mentioning a geometric application of negative Ktheory. This involves the concept, which has been increasingly important in geometric topology during the last several years, of topology with control. For simplicity, we consider one of the simplest illustrations of this idea, as developed in [Pedersen]. Namely, we consider h-cobordisms W between two manifolds M and M’ as in Theorem 2.4.4, but this time with a control map p : W -+ Iw k. The control map p is required to be proper, and its restriction to either M or M’ is required to be surjective. Of course, none of the manifolds W, M, or M’ will be compact. We use the map p to measure “distances,” that is, we define “dist” (2, y) = Ip(s) - p(y)l. Then we require that W have bounded fundamental group, i.e., that there be a fixed constant C such that for every x, y E W, and for every homotopy class of paths from x to y, there be a representative for the class of length < Ip(x) - p(y) I+ C, and similarly that null-homotopic loops be contractible within a set of diameter < C+ the diameter of the loop. The result of [Pedersen] then gives a necessary and sufficient condition for a “bounded” h-cobordism W to have a bounded product structure, in t_erms of an invariant in I?;-k+1(%l(W)), provided that dim W > 5. (Here K_k+l refers to Wh if k = 0 and to K-k+1 if k > 1.) If k = 0, this reduces to the usual s-cobordism theorem (Theorem 2.4.4). Of course, one way a bounded h-cobordism can arise is from a compact h-cobordism W’ with fundamental group 7r x Z’“. The projection of the fundamental group onto Z” induces a map p’ : W’ + Tk, and taking coverings, we get a map p&@&-W-@%@. Theorem 1.7 of [Pedersen] identifies the associated invariant as the image of the original Whitehead torsion in Wh(r x Zk). But there are controlled non-compact problems that do not arise so simply from compact situations. 3.3.6. Exercise. (1) Fill in the details of the argument copied from Theorem 1.3.11, that if S is a ring and if J is an ideal of R contained in rad R, then the map Ko(S) -+ Ko(S/ J) induced by the quotient map S + S/J is injective. (2) Let R be a local ring, not necessarily commutative, and let I = rad R. Assume I” = 0 for some k. Let S = R[tl, t;l, . . . , t,, t,‘],

J = I[tl, t;‘, . . . , t,, t,‘].

Show that S/J is left regular and deduce that NK,(S/ J) = 0 for all n 5 1 and that K-,(,9/J) = 0 for all n > 0. (3) Conclude from (1) that the map Ko(S) + Ko(S/J) is injective. Deduce that NK_,(R) = 0 for n 2 0 and that K_,(R) = 0 for n > 0.

3. Negative K-theory 159

3.3.7. Exercise. Use the results of the last exercise to show that for a finite product of local rings all of whose radicals are nilpotent, all negative K-groups must vanish. Apply this to the ring Z/(m) to complete the calculation of the negative K-groups of (m) C Z for an arbitary positive integer m. 3.3.8. Exercise [KaroubiAlgOp]. Let R be a complex Banach algebra (with unit), and observe that C(S1, R) is also a Banach algebra with pointwise multiplication of functions and with norm

We have an isometric inclusion R q C(S1, R) as constant functions. (1) For all t E S1, the evaluation map at t induces a retraction C(S’, R) --+ R. Show that the induced map on KO is independent of t, hence that Ko(R) sits as a direct summand in Ko(C(S’, R)) in a canonical way. (Use Corollary 1.6.11.) Define K?:(R) = ker (Ko(C(S1, R)) + Ko(R)) . (2) There is a map R[t, t-‘1 w C(S, R) obtained by viewing a Laurent polynomial as a function oft E S1 (identified with the unit circle in the complex plane). This induces a map K-1 (R) -+ KY:(R). Now if R is a Banach algebra, so is M,(R) for all n, so GL(n, R) = (M,(R))X is an open subset of M,(R) by Lemma 1.6.6. It therefore has a natural topology making it a locally contractible topological group. Let GL(n, R)O denote the connected component of the identity in GL(n, R). This contains E(n, R) since each elementary matrix eij (a) is path-connected to the identity via the path eij (ta), 0 < t 5 1. Let GL(R)O = l&GL(n, R)O. Then this is a normal subgroup of GL(R) and the quotient is abelian since GL(R)’ 2 E(R). It is customary to define Kf”p(R) = GL(R)/GL(R)‘. Show (see [Blackadar], Theorem 8.2.2) that there is a functorial isomorphism 8 : KY(R) 3 KFF (R). This is constructed as follows. (a) If u E GL(n, R), then u @ 11-l E E(2n, R) c GL(2n, R)O (Corollary 2.1.3). Choose z E C ([0, 27r], GL(2n, R)O) with z(O) = 12n, z(27r) = u @ u- l. Then let an idempotent p E C(S1, Mz,(R)) = Mzn (C(S1, R)) be defined by p(eit) = z(t)(ln @ O,&(t)-‘. (This is indeed a continuous function of eit, not just a function of t, since z(0) and z(27r) both commute with 1, $ O,.) Define

w4 = [PI - [(ln @ On)].

160

3.

KO and K1 of Categories, Negative K-Theory

Since p(l) = (1, @ 0,),

O([u])

(b)

E

ker (Kc (C(S’, R)) - KO

(I?)) = K?‘?(R).

First show that this is independent of all choices and gives a homomorphism with respect to the block sum operations @ on KY and on KzF. Next, to prove injectivity of 0, suppose 0( [u]) = 0. Stabilizing u if necessary, reduce to the case where p is conjugate to 1, @ 0, in GL(2n), say h(eit)p(t)h(eit)-l

=

(k 0 > n

for some h E C (9, GL(2n, R)). Then show h(ei”)z(t) =

(t> Bz:t)) for some ~1, 22 E C([O, 27r], GL(n, R)) and 0 deduce that u E GL(n, R)O, so that [u] = 0 in KioP(R). Finally, to prove surjectivity of 0, show that every element of K?j’ may be represented in the form [p] - [(ln @ O,)]. (3) Let A = e(s, s-l ), the commutative Banach algebra of absolutely convergent Laurent series (with norm coming from C’(Z)). There is a norm-decreasing homomorphism A + C(S1) with dense image obtained by viewing a Laurent series as a function of u E S1 (identified with the unit circle in the complex plane). Show that K?:(A) % k”(S1 x S’) g Z and that the map K-1 (A) + K!?(A) is surjective. (4 Let R be a complex C*-algebra, that is, a norm-closed subalgebra of the bounded operators L3(‘H) on some complex Hilbert space 3-1 which is invariant under the involution * sending an operator to its adjoint. Then if b E RX, b*b E RX and is strictly positive, so lb1 = (b*b)f E RX (a.r gue as in Lemma 1.6.6) and we have a polar decomposition b = lbju in R with u unitary. In particular, u and u-l each have norm 1. Now if R is a C*-algebra, so is M,(R) for each n. (If R acts on a Hilbert space X’, M,(R) acts on Cm @ E.) Show that if b E GL(n, R), there is a path joining lb1 to the identity in GL(n, R), and hence the class of b in KFP(R) may be represented by the unitary u = IbJ-lb. Then show that with A as above, s H u defines a continuous (in fact norm-decreasing) homomorphism cp : A + M,(R) sending [s], which generates KY(A), to [b] E K?(R). (5) From (2) and (3)ab ove, from commutativity of the diagram 21

K-i(A) - K?;(A) + KE”P(A) ‘p*

1

rp.

1

‘p*

1

K-l(R) - K?:(R) + K ? ( R ) ,

3. Negative K-theory 161

and from the fact that cp can be chosen to have any desired class in KfoP(R) in its image, deduce that the map K-i(R) + K?:(R) is surjective. (6) If X is a compact Hausdorff space, the algebra R = Cc(X) is a C*algebra. (It may be represented, for instance, on a Hilbert space of the form L2(X, ,u), with ,U a measure on X of full support.) Deduce that K-i(R) ++ KU-l(X). This provides many examples of commutative rings with complicated K-1. 3.3.9. Exercise. (Cf. Exercise 3.2.27.) (1) Show from Theorem 3.3.2 that if rr is any group, then

(2) Deduce that the Wall obstruction group Ke vanishes for free abelian groups but is non-zero for 7r x Z if K-1 (Zn) # 0. (An example of a finite abelian group with this property is given in the next Exercise.) 3.3.10. Exercise. Let G and H be finite cyclic groups of orders 2 and 3, respectively, so that G x H is cyclic of order 6. Note that Z(G x H) = ZG @Z ZH. Prom the exact sequences

where w = w, deduce the exact sequences o-*ZG~~~%-~Z(GXH)~ZG~Z~[~]~~, 0+6Z+ZG@z3iZ%332+0, o --+ 22 @Z Z[w] + %G @z Z[w] S z[w] + 0. Note also that there is an exact sequence 0 + 22 @z Z[w] + Z[w] + IF4 + 0. Compute from these and from the fact that Z[w] is a PID that K_1(Z(G x H)) = K_i(6Z) is infinite cyclic and that K_,(Z(G x H)) = 0 for n 2 2. The groups K_n(Z~) have been computed for arbitrary finite groups n by Carter [Carter], and it turns out that K_,(Zr) is always finitely generated for n = 1 and vanishes for n 2 2. Furthermore, torsion can occur in K-1, but it is always of exponent 2.

4 Milnor

‘s PC2

1. Universal central extensions and H2 For the reader who might have been alarmed by the category-theoretic approach of the last chapter, this chapter, which discusses Milnor’s Kz functor, will seem a comforting retreat to more familiar territory. However, we will need to refer to the homology of a group, at least in order to speak of Hz. Since group homology will be needed in a more serious way in the next chapter anyway, we provide a brief introduction to the subject later in this section. The reader who wants a more serious approach to t,he homology theory of groups and its applications should consult a source such as [Brown] or [CartanEilenberg]. First, though, we begin with the theory of universal central extensions, as developed in [Kervaire2] and [Milnor, 551. This a cute and fairly selfcontained topic in group theory, but it’s hard to see at first what it has to do with K-theory. Roughly speaking, the idea here is that K-theory for rings is supposed to measure “abelian” invariants of the highly noncommutative group GL(R). For example, K1 (R) is defined by taking the abelianization of GL(R), in other words, the quotient of this group by its commutator subgroup E(R). Since E(R) is its own commutator subgroup (Proposition 2.1.4), repeating this process with E(R) doesn’t yield anything. However, the deep structure of linear algebra over R should be connected with the deep structure of the group E(R), in other words the relations satisfied by its generators eij(a). One way of measuring these is by looking at extensions of E(R) by abelian groups. There turns out to be a universal such extension St(R), and the (abelian) kernel of the map St(R) ++ E(R) is Milnor’s Kz(R). Even when R is a field, this turns out to be an interesting invariant with lots of numbertheoretic significance. But since the number-theoretic applications of K2 are described quite nicely in [Milnor], we have only touched on the most important of these and have chosen to emphasize some other applications instead.

1. Universal central extensions and Hz

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Universal Central Extensions. 4.1.1. Definition. Let G be any group, and let A be an abelian group. A central extension of G by A is a pair (E, cp), where E is a group containing A as a central subgroup, and cp : E --H G is a surjective homomorphism whose kernel is exactly A. Alternatively, in the language of exact sequences, a central extension of G by A is a short exact sequence

with A central in E. Remark. There are still those who call the above a central extension of A by G, but the terminology above is more in keeping with the formalism of group cohomology, since it turns out that the central extensions of G by A are classified by H2(G, A) ( and everyone agrees that here one should put the G before the A!). 4.1.2. Definition. Next we note that central extensions of G (by arbitrary abelian groups) form a category. If (E, p) and (E’, cp’) are central extensions of (the same group) G, a morphism of central extensions (E, ‘p) + (E’, cp’) is a commutative diagram

A central extension (E, cp) of G by A is called trivial if it is isomorphic in the category of central extensions of G to G x A 3 G, where pl is projection on the first factor. A central extension (E, ‘p) of G is called universal if, for any other central extension (E’, cp’) of G, there is a unique morphism (E, p) -+ (E’, cp’). Not every group has a universal central extension, but if it has one, then it is clear from the definition that any two universal central extensions must be isomorphic (in the category of central extensions of G). 4.1.3. Theorem. A group G has a universal central extension if and only if it is perfect, that is, G = [G, G]. When this is the case, a central extension (E, cp) of G is universal if and only if the following two conditions hold: (i) E is perfect, that is, E = [E, E], and (ii) all central extensions of E are trivial. (Roughly speaking, condition (i) says that E is not too big, and condition (ii) says that it is not too small.)

164 4. Milnor’s KZ

If G is perfect and 14R+F+G+l

is a presentation of G (i.e., a short exact sequence with F a free group), then the universal central extension (E, ‘p) may be constructed as E = [F, F]/[F, R], with ‘p the quotient map [F, F]/[F, R] + [F, F]/R = [F/R, F/R] = [G, G] = G.

Proof. If G is not perfect, it means it has a non-trivial abelian quotient, say A. Let 11, : G --+ A be the quotient map. Now if (E, cp) is a central extension of G, we can construct two distinct morphisms from (E, cp) to the trivial extension G x A 3 G, namely (cp, 1) and (cp, $J o cp) (here 1 : E + A is the trivial map sending everything in E to the identity of A). This shows that (E, ‘p) cannot be universal. Hence, for G to have a universal central extension, G must be perfect. Now assume G is perfect with presentation l+R--+F+G-il. We will show first that any central extension of G satisfying (i) and (ii) is universal. Then to complete the proof, we will show that [F,

FIIP, RI --H IF, FIIR = G

is a central extension satisfying (i) and (ii). This will show in particular that every perfect group has a universal central extension, and since universal central extensions are unique and non-perfect groups do not have universal central extensions, every universal central extension must satisfy (i) and (ii). So assume (E, cp) is a central extension of the perfect group G satisfying (i) and (ii), and let (E’, cp’) be any other central extension of G. Suppose r,!~, $J’ : (E, ‘p) --+ (E’, cp’) are two morphisms of central extensions. For z E E, (p’ o @(cc) = cp(z) = (p’ o $‘(z), hence @(xc) = c~v+!J’(z) for some element c, of the kernel A ’ of ‘p’ : E + G. Similarly, if y E E, then +(y) = c,+‘(y) for some cr, E A’. So 1ct ([xc, Yl) = M(z), ti(Y)l = kzTe4, @J’(Y)1 = w’(~)~1cl’(Y)l = @’ (k? Yl) . (In this calculation we have used the fact that A’ is central in E’.) Hence $ and $’ coincide on commutators. Since E = [E, E] by (i), $J and $’ coincide on all of E. This shows there can be at most one morphism from E to E ’ . We still need to construct a morphism from E to E’. Let E” = E XG E’, that is,

E” = {(z, y) E E x E’ : cp(z) = v’( y)}.

Since cp and ‘p’ are surjective, projection pi on the first factor is a surjective homomorphism E” -H E. The kernel of pl is obviously isomorphic to the

1. Universal central extensions and Hz

165

kernel A’ of cp’, and so is central. So pl : E” + E is a central extension of E. By (ii), this central extension is trivial, which says that there is a homomorphism E -+ E’ commuting with the projections onto G. This means there is a morphism of central extensions from E to E’. Since E’ was arbitrary and we already showed that morphisms from E to E’ are unique, thus E is a universal central extension of G. Finally, let E = [F, F]/[F, R], with cp the quotient map [F, F]/[F, R] + [F, F]/R = [F/R, F/R] = [G, G] = G. To begin with, note that E C El = F/[F, R], which also projects onto G via the quotient map (pi : F/[F, R] + F/R = G, and cp is the restriction of cpi to E. (Since R d F, [F, R] C R, and [F, R] is also normal in F.) Note that the kernel of (pi is contained in R/[F, R], hence commutators of elements of the kernel with elements of El lie in [F, R]/[F, R], which is trivial. Thus the kernel of (~1 is central in El and (E, cp), (El, cpl) are central extensions of G. We need to verify properties (i) and (ii) for E. Directly from the definition of E and El, we see that El has E as its commutator subgroup. On the other hand, since ‘p and (~1 are both surjective onto G, El is generated by E together with the kernel of (~1, which is central. So E = [El, El] = [E . Z(E& E . Z(El)] = [E, E] and E is perfect. (Here, as usual, Z(El) denotes the center of El; the letter 2 comes from the German Zentrum.) This proves (i). As far as (ii) is concerned, let

be any central extension of E. This induces an extension Es = El x GEM 3 El of El, where El XG

E2 = {(G I/> E J% x -J-32 : cpl(~c> = ‘PO$~/)).

This is actually a central extension. Indeed, the kernel of pl : Es -+ El is clearly isomorphic to the kernel of ‘p o $J : E2 -+ G. But since E = [E, E], $([Ez, Ed = [$J(Ez), $(&)I = [E, El = E, and thus & = [Ez, -&I. A. Also, 11, ([Ez, ker cp o $1) C [E, ker ‘p] = 1, so [E2, kercp o $1 C A. Thus for x E ker ‘p o II, and s, t E E2, xsx-1 = s.q and xtx-’ = tz2 for some central z1 and ~2, and x[s, t]z-’ = [xsxF1, xtx-‘] = [sq, tz2] = [s, t]. T h u s x commutes with [Ez, E2]. Since x also commutes with A (A is central), it commutes with all of Ez, and Es is a central extension of El. Since F is free, we can fill in the diagram F -L

E3 = El XGE~ 4 El

166 4. Milnor ’s Kz

to get a homomorphism F + E3 lifting the quotient map F + El. This amounts to a homomorphism 0 : F + EZ such that for IC E F, cp o $ (0 (z)) coincides with the image of x in G Z F/R. So t)(R) G ker cp o + C Z(Ez), and ~([F, R I ) C P(F), e(R)1 C F2, z(E2)l = 1. Hence 9 descends to a map 6 : F/[F, R] = El -+ E2 which, together with the identity map on El, gives a splitting El ---t Es = El xc Es of pi. Restricting to E then gives a trivialization of Y/J : E2 + E, verifying (ii). Thus we have constructed a universal central extension of G. Cl 4.1.4. Remark. Of the conditions in Theorem 4.1.3 for checking when one has a universal central extension, (i) is fairly straightforward, but (ii) is rather difficult to check (without using machinery from group homology theory, which will make it possible to restate the condition in the form Hs(E, Z) = 0). However, the proof of Theorem 4.1.3 gives the following additional piece of information which is sometimes useful. Suppose G is a perfect group and (E’, cp’) is a central extension of G satisfying (i). Then if (E, ‘p ) denotes the universal central extension of G, there is a unique morphism $J of central extensions from E to E’, and $J must map E onto E’ and the abelian group ker cp onto the abelian group ker cp’. Thus condition (i) (without condition (ii)) at least guarantees that one has a quotient of the universal central extension. To see this, note that since E’ = [E’, E’] , to prove surjectivity of $J, it’s enough to show that every commutator is in the image. Let x’, y’ E E’. Then we can choose CC, y E E such that p(x) = (p’(z’), p(y) = cp’(y’), and it follows from the relation cp = ‘p’ o T,!J that x’ = $(x)zi, y’ = $(y)z2 for some ~1, zz E ker ‘p’. Since zi and ~2 are central, 1x’, Y’l =

M(X)%,

ti(Y>Zzl = [@(x)7 ti(Y)l = + ([xc, Yl),

and thus + is surjective. Furthermore, if Z’ E kercp’ and $(z) = z’, then cp’ o $(z) = 1, hence p(z) = 1, which shows that $-‘(ker cp’) C ker cp. The other inclusion is trivial. •I 4.1.5. Examples. Let G and G be connected Hausdorff topological groups and p : 6 + G a continuous surjective homomorphism with discrete kernel D c G. Since D is normal in G, for each d E D, we have a continuous map g t--+ gdg-1 from G to D. Since G is connected and D is discrete, this map must be constant, and thus gdg-1 = d for all d E D, g E 6’, i.e., D is central. So G is a central extension of G. If G is also perfect, which is easy to check in many examples, then it is a quotient of the universal central extension. ff P In particular, let’s take 6’ = SU(2), the group of matrices ( -p B > with a, p E C, IcyI + IpI2 = 1. G will be SO(3), the group of rotations (orthogonal linear transformations of determinant 1) of Euclidean 3-space. G and G are clearly connected and Hausdorff; in fact G is clearly homeomorphic to the unit sphere in C2, or to S3, which is simply connected.

1. Universal central extensions and Hz

167

The Lie algebra of G is defined to be the S-dimensional real vector space fi = {x E Mz(C) : xt = -x, Trx = o}

=

iX -y+iz K

iz -ix > :

y+

I

x, y, z E IR .

We can make this into a Euclidean space via the inner product

(X, Y) = -Tr(XY), since this pairing is clearly symmetric and bilinear, and it’s positive-definite since for X E i, (X, X) = -Tr(XX) = Tr (x . (X”)) 2 0, with equality only if X = 0. Note that G acts on i by conjugation, since ifgEGandXE&

(gXg& = (gXg-‘y = (ijy(Xt)(gt)

= g . (-X) . g-1 = -(gXg_l).

Furthermore, G preserves the inner product on 6 since (gXg-1, gyg-l) = - Tr(gXg-lgyg-1) = - Tr(gXYg_1)

= -Tr(XY) = (X, Y). So we obtain a homomorphism p from G to the connected component of the identity in the orthogonal group of 3, in other words, if we identify fi with lK3, to G. This homomorphism is easily seen to be surjective. Its kernel D consists of matrices g which commute with everything in 5, and thus with everything in

and thus with all of Mz(C). (Any matrix differs by a scalar multiple of the identity from a traceless matrix.) So D = G n {scalar matrices in Mz(C)} = {(;

;), (_d _q)),,i2>

and thus G is a central extension of SO(3) by Z/2. However, G is perfect, since the Implicit Function Theorem shows that the image of the commutator map G x G + G, (g, h) I-+ ghg- ‘he’ , contains an open neighborhood of the identity, and thus generates G. So G is a quotient of the universal central extension of G. There are many other “naturally occurring” pairs (G, G) giving quotients of the universal central extensions of matrix groups. For instance (this example will be important later), let G = SL(n, R) = E(n, R), which

168 4. Milnor’s K2

is a perfect connected matrix group for any n > 2. Then because of polar decomposition, G = SO(n) . expsn, where 5, is the vector space of n x n symmetric matrices with trace zero, and so G has the rotation group SO(n) as a deformation retract. In particular, ri(G) 2 xl(SO(n)) g

Z for n = 2 Z,2 for n > 3

Thus for n 2 3, the universal covering group G of G is a non-trivial double cover, and is again perfect (since the commutator subgroup contains an open neighborhood of the identity, and thus is all of G). Incidentally, one can show that the topological group G cannot itself be realized as a group of real or complex matrices. Thus the universal central extension of SL(W) is non-trivial, and the kernel of this universal central extension has Z/2 as a quotient group. Later we will come back to this from the point of view of J&(W). Homology of Groups. Next we give a quick introduction to the homology of groups, which is an essential tool both for translating the theory of universal central extensions into something computable and for defining and understanding the higher K-groups which will appear in Chapter 5. 4.1.6. Definition. Let G be any group. A (left) G-module A4 is an abelian group equipped with a (left) action of G by automorphisms, satisfying the usual conditions g. (he m) = (gh) . m, 1 . m = m for g, h E G, m E M, or equivalently, a left R-module, where R = ZG is the group ring of G. Note that any such M may also be made into a right G-module if we define m : g =+f g-l .m, though there is one source of possible confusion: if G is abelian, then R is a commutative ring, and we customarily make any left module for such a ring into a right module by defining m . I- =,+f r . m, and this convention disagrees with the previous one. When it makes a difference, we will specify which right action we are using. We will denote by Z the trivial G-module which is Z as an abelian group and with g . n = n for all g E G, n E Z. (In this case, when G is abelian, the two right actions agree.) 4.1.7. Definition. Let G be any group, M a G-module. We define the homology and cohomology groups of G with coefficients in M as follows. First note that we can construct a free resolution of the trivial Gmodule Z by letting Pj = ZGj+‘, the free abelian group on the (j + 1)-st Cartesian product of G with itself, for j 2 0, and defining boundary maps dj : Pj + Pj-1, j 2 1, by

k=O

where .

(Note that the right action used here is the same as the left action if G is abelian, and is in general not the same as the “flipped left action” m-g = g-1. m.) The homology groups of G with coefficients in A4 are the homology groups of the complex

C.(G, M) = P. '&G M 2

ZZ’ @z M,

and the cohomology groups of G with coefficients in M are the homology groups of the dual complex

c’(G, M) = HomzG(P.,

M) g HomzW’, M)a

(Here we are always using the left module structure on M. TO identify Pj @ZG M with ZGj @Z M, we view Pj as the free right ZG-module OP generators (1, gr, . . . , gj). To identify HomzG(Pj, M) with Homz(ZGJ, M), we let F E HomzG(Pj, M) correspond to f E Homz(ZGj, M), where

fk.71, -**> gj> =F(l, F(go, . . . > Sj) =

90 ’

91, glg2, ***T f(L7cj-1fJ1,

glg2..*Sn),

9T192, . . . T

ST2iSj) -1

The lowest-dimensional parts of these complexes are of particular interest, so we write them out explicitly:

170 4. Milnor’s K2 C.(G, M)

. . . i?+ ZG2 gz M ++ ZG ,& M dl:g@mHg’m-m, M,

:

& ((91, . *. , h) @ m) =

(92&l, . . . , 57dT1) ~8 91 . m

+ &qgl,

. . .) 6.. .) 4 @m,

j=l C.cG, Mj :

M dO:dOm(g)=g~m--m

F

HomZ(ZG, M)

2 Homz(ZG2, M) z . . . ,

d”f (go, . . ., Sn> = 90 . fbl, . . . , gn) + &~~f(~o, . . ..gj-1gj...., gn) j=l + (-l)“+‘f (go, . . . , a-1).

Note that when the G-action on M is trivial, dl = 0 and do E 0, so Ho(G, M) = H’( G, M) = M. Also in this case, we have & ((91, 92) @ m) = (g2gF1 - g2 + gr) @ m, so Hr(G, M) = ZG 63~ M/ im d2 = G,b @Z M, while H1 (G, M) is simply Hom(G, M). More generally, when the action of G on M can be arbitrary, we see that He(G, M) = MG, the quotient of M by the submodule generated by the elements g. m - m, g E G, m E M. Similarly, H’(G, M) = MG, the elements of M left fixed by all elements of G. The 1-cocycles f E Z1 (G, M) are functions f : G -+ M such that 90 . f(a) - f(gm) + f(d = 0 for all go, gr E G, or such that f(gsgr) = f(ge) + go . f(gr). The lcoboundaries are those of the special sort f(g) = g.m-m for some m E M; H1 (G, M) is the quotient of Z’(G, M) by this subgroup. Similarly, 2cocycles f E Z2(G, M) are functions f : G x G + M such that go . fh, s4 - fkwl, 92) + fko, cm) - fh 91) = 0 for a~1 go,gl, a E G. While we will not develop that much of the homology theory of groups, we present at least a few tools for computing homology and cohomology and develop the relationship between Hz and H2 and the theory of central extensions. The first basic facts are contained in the following proposition. 4.1.9. Proposition. Let G be a group. For each k 2 0, M u) Hk(G, M) and M -+P Hk(G, M) are (covariant) functors on the category of G-modules. If M is a projective G-module, then Hk (G, M) = 0 for all k > 0. Similarly, if M is an injective G-module, then H”( G, M) = 0 for all lc > 0. If 0 + MI % M2 -% M3 + 0 is a short exact sequence of G modules, there are associated long exact sequences

. . . 5 fb+l(G,

M3) -% Hk(G, Ml) a -&(G, M2) k Hk(G,

M3) 3 Hk_l(G, MI) -% . . .

1. Universal central extensions and

Hz

171

and

. . . -% Hk-l(G, MS) -% Hk(G, Ml) = H”( G, Mz)

-% H”(G, M3)

-fz, Hkl(G, Ml) =. . . .

Proof. It is obvious from the definition that homology and cohomology are functorial. If M is a free ZG-module, with a free basis indexed by a set I, then for any right ZG-module N, N @zG M is naturally isomorphic to N’, and so tensoring with M preserves exactness. A similar argument applies if M is a direct summand in a free module. Thus, if M is projective, since the complex P. is exact, so is P. @zG M, and so Hk (G, M) = 0 for all k > 0. Similarly, if M is injective, then HomzG(., M) preserves exactness and so HomzG(P,, M) is exact, so that Hk(G, M) = 0 for all k > 0. The statement about long exact sequences follows immediately from Theorem 1.7.6 (the Fundamental Theorem of Homological Algebra), since a short exact sequence of G-modules yields short exact sequences

and 0 + HOmZG(P.,

Mr) a Homzc(P., Mz) L Homzc(P.7

M3) + 0

of chain complexes. 0 4.1.10. Corollary. If G is a group and M is a G-module, then homology of M can be computed from a projective resolution of M, while cohomology can be computed from an injective resolution. More precisely, if

is exact and each Nj is G-projective, then H.(G, M) is the homology of the complex z @zG N.. Similarly, if

is exact and each Nj is G-injective, then H’( G, M) is the homology of the complex N.” . Proof. This follows by iteration, splitting the resolution into a series of short exact sequences and using Proposition 4.1.9 over and over again. For instance, consider the case of a projective resolution N. of M. First consider the short exact sequence

0 -+ Ni/(im&) 5 No + M -+O.

172 4. Milnor’s K2

From this we obtain a long exact sequence of homology groups, but since Nc is projective, Hk(G, No) = 0 for k > 0. Thus Hk(G, M) g Hk-r(G, Nr/(im&)) for k > 2, and similarly there is an exact sequence 0 --f H1(G, M) -, Ho(G, Nl/(im&))

ti Ho(G, No) --t Ho(G, M) +

0.

On the other hand, we have an exact sequence 0 + N2/(imds) 3 Nr --+ Nr/(im&) --+ 0,

(4.1.11)

and Nr is projective. Repeating the argument, Hk(G, M) g &--l(G, Nl/(im&)) g

Hk-2(G, N2/(imd3))

for k 2 3, and we obtain an exact sequence 0 -+ H2(G, M) g Hr(G, Nr/(im&)) + Ho(G, =

WGm&))

Ho(G, Nr) + Ho(G, Nr/(im&)) + 0.

Putting this together with (4.1.11), we see that HI(G, M) and Ho(G, M) are the lowest-degree homology groups of the complex Ho(G, IV.) g Z@ZG IV.. Then we continue inductively to compute H2(G, M), and so on. 0 For future applications, the following easy consequence of Corollary 4.1.10 is often useful. 4.1.12. Corollary (“S hapiro’s Lemma ”) . Let G be a group and let H c G be a subgroup of G. Let M be an H-module. Then there are natural isomorphisms Hj(G, ZG 63’~~ M) 2 Hj(H, M) for all j. Proof. Choose a ZH-projective resolution N. of M. By Corollary 4.1.10, H.(H, M) is the homology of the complex Z I&H N.. However, if gi is a set of representatives for the right H-cosets in G, then ZG is a free right ZH-module with basis gi, so ZG @ZH N. = Bi gi @ N. is a ZG-projective resolution of ZG @.ZH M. Thus H.(G, ZG C?QH M) is the homology of the complex Z&,GZG@ZHJJ.~Z@ZHN.,

which proves the result.

0

For purposes of studying central extensions, we will be particularly interested in homology and cohomology of G-modules with trivial G-action. These are related by the following.

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173

4.1.13. Theorem (“U niversal Coefficient Theorem”) . Let G be a group and let M be an abelian group, viewed as a G-module with trivial G-action. Then there are short exact sequences 0 + Ext; (l&-r (G, Z), M) -+ H”( G, M) -+ Homz (HI, (G, Z), M) + 0 for all k, which split (though not in a natural way). In particular, Hz (G, M) = 0 for all G-modules M with trivial G-action if and only if the abelianization of G is free (abelian) and HZ (G, Z) = 0. Proof. Recall that H. = H. (G, Z) was defined to be the homology of P. ~IZG Z. However, since P. is a G-projective resolution of Z, by Corollary 4.1.10 we can also compute it as the homology of the complex C. = Z @zG P. z ZG’. On the other hand, H’( G, M) is the homology of Homzc(P,, M) E HomZ(ZG’, M) Z Homz(C., M) with the dual differential. Let Zk = ker(Ck 3 Ck-1) and Bk-_l = im(Ck 5 C&r), so that Hk(G, Z) = Zk/Bk. Note that Ck iS a free abelian group; hence its subgroups Zk and Bk are also free. We have short exact sequences 0 --i Z, --i Cr, 3 Bk-r -+ 0 and

0 -+ Bk --) Z, --) HI, + 0.

Since Bk_-l is free abelian, the first of these splits, and functor HomZ(., M) to a (split) short exact sequence

SO

goes under the

d;-, 0 + Homz(Bk_r, M) - Homz(Ck, M) -+ Homz(Zk, M) + 0. (4.1.14) Exactness of the second is not preserved in general, but from the definition of Ext there is an exact sequence 0 -+ Homz(Bk, M) + Homz(Zk, M) + Homz(Bk, M) 4 Exti(Hk, M) -+ 0. (4.1.15) Now note that (4.1.14) may be viewed as giving a short exact sequence of chain complexes 0 + Homz(B._r, M) + Homz(C., M) + Homz(Z., M) --+ 0, where the outside chain complexes have vanishing differentials. So from the Fundamental Theorem of Homological Algebra, we obtain a long exact homology sequence . . . --+Homz(Zk-1, W +Homz(&-1, M) +

H”(G, M) + Homz(Zk, M) --+ Homz(Bk, M) -+ *.. .

174 4. Milnor ’s K2

Substituting from (4.1.15) we obtain the desired short exact sequence 0 + Ext; (H/+-l (G, Z) , M) + H”(G, M) -+ Homz (Hk (G, Z), M) -+ 0. The sequence splits since (4.1.14) splits (though not naturally). To prove the last statement, recall that Hl(G, Z) is the abelianization of G. Hence Exti (HI (G, Z) , M) vanishes for all M if and only if HI (G, Z) is Z-projective, i.e., free abelian. Similarly, Homz (Hz (G, Z) , M) vanishes for all M if and only if Hz (G, Z) vanishes. 0 Now it is time to make the connection between group homology and the theory of extensions. This follows from the following basic classification theorem, due originally to Eilenberg and Mac Lane. The theorem has a version for non-central extension extensions and even a version for extensions by a non-abelian normal subgroup, but in the interests of simplicity we stick with the simplest case, which is all we will need for applications to K-theory. 4.1.16. Theorem. Let G be a group and let A be an abelian group. Then the isomorphism classes of parameterized central extensions of G by A, that is, triples (E, ‘p, L), where (E, ‘p) is a central extension of G and L : A + E is an isomorphism of A with ker cp, naturally form an abelian group Ext(G, A>, in which the trivial extension gives the O-element. This group is naturally isomorphic to H2(G, A), where we view A as a trivial G-module. Hence every central extension of G by A is trivial if and only if H2(G, A) = 0. Proof. We should make precise what we mean by isomorphism; two triples (E, cp, L) and (E’, cp’, L’) are isomorphic if and only if there is an isomorphism of central extensions from E to E’ compatible with L and L’, i.e., a commutative diagram with exact rows l-----+A”E&G-1

LI

,

l-A-E’AG-11. Next, we explain the group structure on Ext(G, A). If (El, (~1, ~1) and (E2, (~2, ~2) are central extensions of G by A, we define their Baer sum as follows. First let E = El xG E2, i.e., ((2, y) E El x E2 : cpl(z) = cps(y)}. Note that this group comes with a natural surjection cp : E ++ G given by cp(z, y) = cpl(z) = 92(y). The kernel of cp is central, but it’s too big; it’s isomorphic to A x A. Therefore we define (4.1.17)

E3 = E/ ((~1 (u) , -LZ (a)) :

a E

A}.

Note that cp factors through Es and gives a map ‘p3 : E3 --H G. The kernel of (~3 is central and is given by {(h(m) 7 L2 (a211 :~l,a2~A}/{(~l(a),-~2(a)):a~A),

1. Universal central extensions and Hz

175

so we can define an isomorphism ~~ : A + ker ‘p3 by L3(U) = [(h (a) 7 L2 (O>>l = [CL1 (0) 7 L2

(Here we use the relation that [(LI (-a),

~2 (u))] =

(a>>1 *

0.) We define

[(El, ‘pi, 41 + I(E2, (~2, ~211 = ICE37 93, ~3)1$ This addition operation is actually commutative, since if we define (Ed, (~4, ~4) similarly but with El and Es interchanged, then we have a commutative diagram

1----+A4e,E4&G--+1 with $J defined by the “flip.” It is easy to verify that the Baer sum is associative on isomorphism classes of central extensions, that the isomorphism class of the trivial extension (G x A

P I,

i2),

where pl is projection on the first factor and i2 is injection into the second factor, acts as an identity with respect to the Baer sum, and that the class of (E, cp, L) has as its inverse (E, cp, -L). Thus we obtain an abelian group Ext(G, A) of parameterized central extensions of G by A, in which the trivial extension gives the O-element. Now we want to set up an isomorphism between Ext(G, A) and H2(G, A). First we define a map Q : Ext(G, A) --+ H2(G, A), then we’ll define the inverse map 9 : H2(G, A) + Ext(G, A). Start with a class in Ext(G, A) represented by (E, cp, L). Choose a (set-theoretic) section s : G -+ E, that is, a map such that cp o s = idG. We may suppose that s(1~) = 1~. For g, h E G, cp(s(gh)) = cp(s(g))cp(s(h)), so we can define a map f : GxG + A by s(gh) = L 0 f(g> h)s(gb(h). Since 1~ = s(lc), f(g, 1~) = f(l~, 9) = 0 ad f(s, 9-l) = f(g-I, 9) for all g E G. If g, h, k E G, then s(gh)s(k) = LO f(g, h)s(g)s(h)s(k), w h i l e s(g)s(hk) = LO f(h, k)s(g)s(h)s(k), so the associative rule gives L0

fh hk)L 0 f(h ~)s(g)s(h)s(k) = L 0 f(g, hk)s(g)s(hk) = s(ghk) = L 0 fkh kb(gh)s(k) = L 0 f(gh I~)L 0 f(g, h)s(gb(hb(k),

or f(g, hk) + f(h, lc) = f(gh, k) + f(g, h). (We write products multiplicatively in E but additively in A.) Comparison with the formulas in Definition 4.1.7 shows that this is precisely the condition for f to define a 2cocycle in Z2(G, A). So we let @([(E, cp, L)]) = [f] E H2(G, A). Of course, it is not immediately obviously that this is independent of the choice of s.

176 4. Milnor’s Kz

But if s’ is some other choice for s, then we must have s’(g) = s(g)L o u(g) for some map u : G -+ A. Then if f’ is the 2-cocycle defined by s, we have

S(&>L 0 u(gh) = s’(gh) = L 0 f’(g, h)s’(g)s’(h) =LO

f’k, h)L O ‘1L(S)L 0 ‘Ll(h)s(g)s(h),

and comparison with the definition of

f gives

fb, h) + 4gh) = f’h h) + 4s) + U(h)> which says that f’ differs from f by the coboundary of IL. Hence f’ and f define the same cohomology class and @ is well defined. To show that @ is a homomorphism, note first that @ sends the Oelement of Ext(G, A) to the O-element of H2(G, A), since when (E, ‘p, L) = (G x A, pl, i2), we can take s = ir, which gives f = 0. Next we show that @ respects the Baer sum. Given (El, (~1, ~1) and (Es, 972, ~a), choose corresponding sections sr and ss giving cocycles fl and fi, and let E = El xG E2 and E3 be as in (4.1.17). Note that s = (sr, sz) gives a section of (E, ‘p) which descends to a section s3 for Es. Then if g, h E G, we have

s&h) = [(Sl(gh), s&h))1

= [(sl(s)sl(hL)~l

h), sdghbz 0 fib, h))l fdg, h))l = sdgh(h)~3 (fi(g, h) + fdg, h)) 7 0 fib

= sdgbdh)K~lofi(g,

h), ~20

so that the cocycle f3 defined by ss is just fi + fi. This shows that @ is additive. To complete the proof, we show that @ is bijective. First suppose @([(ET P, ~)l) = 0. Th is means that if we choose a section f as above, the corresponding cocycle f is the coboundary of some u : G --+ A, i.e.,

f (g> h) = 4gh) - u(g) - u(h),

g, h E G.

Replacing s by s’, where s’(g) = s(g)(L o u(g))-l, we have s’(gh) = s(gh)(L o u(gh))-’ = s(g)s(h)l 0

(f (g, h) - u(gh)) = 4gMh)l. (4gh) - u(g) - u(h) - u(gh)) = s’(g)s’W, so s’ is a homomorphism, which shows (E, cp, L) is trivial. Thus Cp is injective. Now we construct a right inverse 9 for @, showing that Q is surjective. Let f E Z2(G, A), which we can take (by changing it within its cohomology class) to be normalized so that f = 0 on {lo} x G and on G x (1~) (this

1. Universal central extensions and Hz

177

implies f(g, g-l)) = f(g-‘, g)) and let E = G x A as a set, but with the following binary operation: (91, aI> . (92, a2> = (91 . Q2, a1 + a2 - f(g1, 92)).

Since f is normalized, (lo, 0~) acts as an identity element with respect to the operation . . Furthermore, we have from the cocycle identity ((91, 4 . (92, a2)) . (93, a3) = (91 . Q2, a1 + a2 - f(g1, 92)) . (93, a3) = (91 . Q2 . Q3, a1 + a2 + a3 - f(g1, 92) - f(s1 . Q2, 93)) = (91 * 92 . Q3, a1 + a2 + a3 - fb2, 93) - f(g1, Q2 .93)) = (L71, al> . (92 . Q3, a2 + a3 - f(g2, 93)) = (91, 4 . ((92, a2) . (93, a3)).

Also, (9, o)

’ (g-? -a + fk?, 6)) = (1G, OA),

(g-l, --a + fb, g-‘>> ’ (9, a) = (1G, OA), so E is a group with respect to . . Define ‘p : E + G to be projection on the first factor and define L : A + E by L(U) = (lo, u). Then it is clear that cp and L are homomorphisms and that L(A) is central in E and equal to the kernel of cp. Furthermore, s : G + E defined by s(g) = (g, 0) is a section which gives rise to the cocycle f since

s(gh) = (gh7 0) = (9, 0) * (4 fkJ> h)) = 4gb(~Mfk7, h)), TV) $1 - KEY 97 L)l is a right inverse to a, showing that + is bijec4.1.18. Corollary. If G is a perfect group, then a central extension (E, ‘p ) of G is universal if and only if Hl(E, Z) = 0 and Hs(E, Z) = 0. Proof. This follows immediately from Theorems 4.1.3,4.1.13, and 4.1.16. (Note that a group is perfect if and only if its Hi vanishes.) q In fact, one can go a bit further. 4.1.19. Theorem. Let G be a perfect group. Then the kernel of the universal central extension (E, ‘p) of G is naturally isomorphic to A = Hz (G, Z), and under the isomorphisms Ext(G, A) g H2(G, A) cs Homz (HZ (G, Z), A) defined by Theorems 4.1.16 and 4.1.13, the class of (E, ‘p ) corresponds to the identity map Hs(G, Z) -+ A. (The Ext term vanishes since H1 vanishes.) The proof of this theorem requires developing some of the theory of how homology and cohomology behave under group extensions. To do this in the greatest generality requires the theory of spectral sequences and would take us a bit too far afield in homological algebra. However the following special case of the theory can be done directly.

178 4. Milnor ’s Kz

4.1.20. Theorem (“I nflation-Restriction Sequence”) . Let G be a group, N a normal subgroup, and A a G-module. Then there is a natural exact sequence 0 + @(G/N, AN) % H1(G, A) = H1(N, A)G/N 2 H’(G/N, AN) % H2(G, A). Here “res” comes from restriction of cocycles from G to N and “inf” denotes inflation, composition of cocycles on G/N with the quotient map G -% G/N. The action of G/N on HI (N, A) is induced from the conjugation action of G on N and from the action of G on A. If G acts trivially on A and N is central in G, the exact sequence simplifies to 0 4 Hl(G/N, A) % H’(G, A) = H1(N, A) 2 H2(G/N, A) % H’(G, A), where the map H1(N, A) S Hom(N, A) 5 H2(G/N, A) sends u : N + A to the class of uof, where f E Z2(G/N, A) is a cocycle defining the central extension of G/N by N as in Theorem 4.1.16. Proof. Note that AN is a G/N-module, since if 4 E G/N, a E AN and q(g) = jr, then jr . a =def g. a will not change if we replace g by gn, n E N. Ifu:G/N+A N is a 1-cocycle, then u o q : G -+ AN is a 1-cocycle, since for g, h E G, u o q(gh) = ‘IL(&) = u(b) + 4. u(k) = u o q(g) + g . (u o q(h)). Furthermore, if uo q is the coboundary of some a E A, then u(ti) = g. a - a, hence taking g E N shows that a E AN and u is the coboundary of a. Thus Hl(G/N, AN) -% H1(G, A) is injective. It is clear that there is a homomorphism H1(G, A) z H1(N, A). To show that the image is fixed by G/N, consider a 1-cocycle u : G --) A and let g E G, ti = q(g), n E N. Then (9. @‘u))(n) =def 9. u(g-‘ng) = 9. (u(g-l) + 9-l . u(ng)> zz g . (-9-I . u(g)) + 9.9-l . u(ng) =

-u(g) + u(ng) = u(n) + n. u(g) - u(g).

Thus g. (res U) differs from res u by the coboundary of u(g), and so jr fixes the cohomology class of resu. That res o inf : H1 (G/N, AN) + H1(N, A) is 0 is trivial. We must show that if a 1-cocycle u : G -+ A restricts on N to the coboundary of some a E A, then u is up to a coboundary the inflation of a cocycle on G/N with values in AN. Replacing u by u - 6a, we may suppose u vanishes on N. Then for g E G and n E N, u(gn) = u(g)+g.u(n) = u(g)+0 = u(g), so u is

1. Universal central extensions and H2

179

constant on cosets of N. Furthermore, u takes values in AN, since we then have (by normality of N) u(g) = u(ng) = u(n) + 7~. u(g) = 0 + n. u(g) = n. u(g), so u is the inflation of a cocycle G/N + AN. We prove the last part of the theorem only in the case where A is a trivial G-module, which is the only case we’ll need for applications. The general case works the same way but the calculations involved are much messier. We define the map a : H1(N, A)GIN --+ H2(G/N, A). Let u : N --+ A be a 1-cocycle, i.e., a homomorphism, which is invariant under conjugation by elements of G. (Since A is abelian, u is automatically fixed under conjugation by elements of N.) Fix a section s : G/N + G with S(~GIN) = 1~. We define $ : G/N x G/N --t A b y q!+j, k) = U(s(Q-is(g)%(giL)). The same calculation as in Theorem 4.1.16 shows that 1c, is a 2-cocycle with values in A, and that its cohomology class doesn’t depend on the choice of s. We define a( [u]) to be the class of +cI; it is obvious that d is a homomorphism of abelian groups. Note that in the important special case when N is central, the condition that u be G-invariant is vacuous, and 1c, is just u of, where f : G/N x G/N + N is the cocycle determined by s and the central extension of G/N by N, so in this case d : H1 (N, A) -+ H2(G/N, A) is just composition with the class of f. It is also clear in general that II, vanishes identically if s is a homomorphism. Let’s show next that d vanishes on the image of res. Suppose u : G --) A is a homomorphism; we denote its restriction to N by the same letter. Define a map v : G/N -+ A by v(g) = u(s(g)). Then 6v(g, A) =&f v(g) + v(iL) - ‘u(giL) = 21(Q)) + u(s(iz)) - u(s(&)) = -$J(& iL), so q5 is a coboundary and d(res u) = 0. In the other direction, if u : N -+ A is a homomorphism fixed under conjugation by elements of G and a(u) = 0, then $J as defined above is a coboundary, say of some -v : G/N -+ A. In other words, U(s(~)-is(g)-ls(g~)) = -v(g) - zl(tL) + v(&). We may suppose w(~G/N) = 0. Let ti(g) = v(g) + u(s(g)-ig). T h e n it is clear that ii agrees with u on N and that for g E G and n E N, ii = ii(g) + u(n). Finally we have ?qgh) = v(&) + zA(s(giL)-lgh) = v(g) + v(h) + U(s(~)-is(g)-is(gq) + u(s(g&igh) = w(g) + v(h) + U(s(lL)-‘s(g)-lgh) = G(h) + v(g) + U(h-‘s(g)-igq = C(h) + C(g),

180 4. Milnor’s K2

since u is invariant under conjugation by h, and thus fi is a homomorphism extending u and u = res C. It remains to prove exactness at H2(G/N, A). First of all, inf od = 0, for if u : N + A is a homomorphism fixed under conjugation by elements of G and $J is defined as above, then inf $(g, h) =&f u(~(iL)-~s(~)-~s(&)) = u(s(h)-‘h) + u(h-ls(cj)-ls(&)) = u(s(iL)-‘h) + u(s(cj)-‘s(&)h-l) = u(s(h)-‘h) + u(s(4)-‘9) + u(g%(&)h-‘) = u(s(k)-‘h) + u(s(9)-lg) + u(h-1g-1s(9iL)) = bv(9, h), where w(g) = u(s(9)-‘9). Finally, suppose 1c, E Z2(G/N, A), which we may suppose is normalized to vanish when one of its arguments is ~GIN, and inf 1c, is a coboundary, say 6v with v : G --f A. We need to show that the class of $J is in the image of a. What we are given translates into the condition +(9, h) = v(9) + v(h) - u(9h). Since the left-hand side vanishes when g or h lies in N, this says in particular that v restricted to N is a homomorphism, and that v(gn) = u(n9) = v(9) + u(n) for n E N. Thus the restriction of w defines a class in H1(N, A) G/N. The class d(v]N) is defined by the cocycle $J’(9, it) = v(s(~)-rs(9)-‘s(9i)). Then

(Ic, - $J’>(jr,

iL) = v(s(ti)) + ‘u(s(jL)) - v(s(9)s(h)) - v(s()L)-l&j)-ls($)).

Since n = s(h)-‘s(9)-‘s(9~) E N, the identity v(s(jr)s(h)) + v(n) = v(s(9)s(iL)n) gives (Ic, - @)(9, A) = u(a(9)) + v(s(Q) - w(s(9lL)) = 6(v 0 s)(9, jL), and [$I = d(v]N), as desired.

Cl

Proof of Theorem 4.1.19. Let G be a perfect group, let (E, cp) be a perfect central extension of G with kernel ker cp = K, and let A be a trivial E-module. The exact sequence of Theorem 4.1.20 becomes 0 = H’(E, A) =+ H1(K, A) 2 H2(G, A) 3 H2(E, A). Now for (E, ‘p) to be the universal central extension of G, E must be perfect and H2(E, A) must vanish for every A. So this can happen only

1. Universal central extensions and H2

181

if H1(K, A) = Homz(K, A) 3 H2(G, A) is an isomorphism for every A. But by Theorem 4.1.13, H2(G, A) “= Homa (HZ (G, Z) , A), so we must have Homz(K, A) E Homz (HZ (G, Z) , A) for every abelian group A, which is only possible if K c H2 (G, Z). F’urthermore, from the description of a in Theorem 4.1.20, we see that the P-cohomology class of the extension of G by K must correspond to an isomorphism Hz (G, Z) --+ K, which we can take to be the identity after reparameterizing K . Cl Remark. Milnor in [Milnor] gives a different proof of Theorem 4.1.19, by identifying the kernel of the map [F, F]/[F, R] + [F, F]/R in Theorem 4.1.3 directly with H2(G, Z). This comes from applying the analogue in homology of Theorem 4.1.20 to the group extension

This basically concludes the discussion of the relationship between central extensions and homology. However, for future use in studying the homology of groups such as SL and GL, we mention a few more basic facts about group homology. 4.1.21. Definition. If H -% G is a homomorphism of groups and A is a G-module (viewed also as an H-module via a), there are induced maps (Y* : H.(H, A) -+ H.(G, A). (Merely take the complex defining H. (H, A) and apply cr to each copy of H.) When (Y is the inclusion of a subgroup, this map is commonly called corestriction, since it is the analogue in homology of H’( G, A) z H’ (H, A). However, when H is of finite index T = [G : H] in G, there is also a natural map in the other direction, called the transfer, sometimes denoted Tr$ or (Y!. This may be defined as follows. Note that ZG is naturally the free ZH-module on the set H\G, so that if A is a free ZG-module, Ho(H, A) 2 Ho(G, A)H\G. (This uses the fact that H\G is a finite set, since in general we only get a direct sum of copies of HO (G, A), not a direct product.) Thus there is a diagonal T times map Ho(G, A) + Ho(H, A) Z Ho(G, A)H\G, given by a H (a), called the transfer. In general, we can resolve A by free ZG-modules, use Corollary 4.1.10, and do this to each step of the resolution. There is another equivalent way of defining corestriction which is sometimes useful. Namely, we can use Shapiro’s Lemma (Corollary 4.1.12), which sets up a natural isomorphism Hj(H, A) 2 Hj(G, ZG @zH A). The map (Y* is easily seen to be the composition of this isomorphism with the map Hj(G, ZG ‘&H A) + H.(G, A) induced by the map of G-modules ZG @&H A + ZG @zG A E A coming from the fact that A is a G-module and not just an H-module.

182 4. Milnor ’s KS

4.1.22. Proposition. If cx : H L) G is the inclusion of a subgroup of finite index r = [G : H] and A is a G-module, then CY, o cr! is multiplication by r on H.(G, A). Proof. If A is free, then Ho(H, A) E Ho(G, A)H\G and T times

T times

a*OcxycL)=a*(~)=h=m.

In general, we can resolve A by free ZG-modules, use Corollary 4.1.10, and apply this at each step of the resolution. Cl 4.1.23. Theorem. If G is a finite group of order r and A is a G-module, then Hj (G, A) is a group of exponent r for each j > 0, and Hj (G, Z) is a finite abelian group of exponent r for each j > 0. Proof Let H be the trivial one-element subgroup of G. Then Ho(H, A) = A and Hj(H, A) = 0 for j > 0 (this is obvious from Definition 4.1.7). Applying Proposition 4.1.22 to the inclusion (Y of H into G, we see that oI* o Q! is multiplication by r on Hj (G, A), while of course this composite is zero for j > 0. So multiplication by r on Hj(G, A) acts by zero for j > 0. This proves the first statement. Furthermore, the abelian groups Pj in (4.1.8) are finitely generated if G is finite. So each Hj(G, A) is finitely generated if A is finitely generated, in particular if A = Z. Since a finitely generated abelian group of finite exponent is finite, the last statement follows. 0 4.1.24. Corollary. Let G be a finite group, let p be a prime dividing the order of G, and let Pp be a Sylow p-subgroup of G. Then for each j > 0, the natural map Hj(P,, Z) + Hj(G, Z) is surjective onto the p-primary part. In particular, if Hj(P,, Z) = 0, then Hj(G, Z) has no p-torsion, and if Hj(P,, 9) = 0 for each p dividing the order of G, then Hj(G, Z) = 0. Proof. Apply Proposition 4.1.22 to the inclusion (Y : Pp - G. Thus CK, o cr! is multiplication by [G : Pp], which is relatively prime to p, and is thus an isomorphism on the pprimary part of Hj(G, A) for j > 0. So a, is surjective on the p-primary part. Cl 4.1.25. Exercise. Let G be a cyclic group of finite order r, and identify ZG with Z[t]/(P - l), where t is a generator of G. Show that

where N is multiplication by 1 + t + . . . + t”-l, gives a free resolution of the trivial G-module %. Deduce that Hj(G, Z) = 0 for j positive and even, and that Hj(G, Z) g G for j positive and odd. Show also that Hj(G, Z) = 0 for j positive and odd, and that Hi(G, Z) Z G for j positive and even. The generator of H2(G, Z) corresponds to the non-trivial central extension 0 + Z r, Z + Z/r -+ 0.

1. Universal central extensions and Hz

183

4.1.26. Exercise. Let V be the Klein 4group, the subgroup

((B

i

;),

(;

;1

;);

(;

;

!l).(s

J61

!I)}

of SO(3). (The “V” stands for Vieng-uppe, German for “4group.“)

(1)

Show by direct calculation that Hz(V, Z) %’ Z/2, and deduce from Theorem 4.1.13 that P(V, Z/2) 2 (Z/2)3.

Determine which elements of this group correspond to the various groups of order 8 which are central extensions of V by Z/2. (2) Examples 4.1.5 may be interpreted as exhibiting Z/2 as a quotient of Hs(SO(3), Z). Show that the inclusion V L) SO(3) induces a non-zero map Z/2 s Hs (V, Z) + Hz (SO(3), Z) which is a splitting for this Z/2 factor, by noticing that the inverse image V of V in SU(2) is the quaternion group Q of order 8. (Since Q is nonabelian, this means the central extension of SO(3) is non-trivial on the subgroup V.) (3) Show that Hi(Q, Z) Z V (this is almost trivial) and that Hz(Q, Z) = 0 (it helps to construct a suitable resolution). Thus the quotient map Q + V induces an isomorphism on Hi but is not surjective on Hz. 4.1.2’7. Exercise. This exercise will exhibit an interesting finite example of a universal central extension. (1) Let G be the subgroup of SO(3) consisting of rotations mapping a regular icosahedron to itself. Since G acts transitively on the 20 faces of the icosahedron, and each face is an equilateral triangle, and there are clearly 3 rotations stabilizing each face (the identity and rotations by 2~13 in either direction about an axis through the center of the face), G is a group of order 60. It is clear that the stabilizer Sf of a face is a Sylow S-subgroup and that the stabilizer S, of a vertex is a Sylow 5-subgroup. (2) Show that one may position the icosahedron so that two of the edges are parallel to each of the three coordinate axes in R3. De duce (by looking at the effect of rotations by r around these axes) that G contains the group V 2 Z/2 x Z/2 of Exercise 4.1.26 as a Sylow 2-subgroup. (3) From Exercise 4.1.25, Hz(Sf, Z) = 0 and Hz(&, Z) = 0. From Exercise 4.1.26, Hz(V, Z) 2 Z/2. Deduce from Corollary 4.1.24 that Hg(G, Z) has order at most 2. (4) Show that G is perfect. In fact it is isomorphic to the simple group As, and is the smallest non-trivial perfect group. (One way to do this is to divide the 30 edges of the icosahedron into 5 equivalence

184

4.

Milnor’s KZ

classes of 6 edges each, where each equivalence class consists of the edges pointing in directions parallel or perpendicular to the direction of a given edge. Then you just need to show that G acts faithfully, i.e., without kernel, by permutations of the 5 equivalence classes. Since A5 is the only subgroup of S’s of order 60, this shows G is isomorphic to the simple group As.) (5) From (2) and (3), deduce that H2(G, Z/2) has order at most two, and that either Hs(G, Z) = 0 or else G has exactly one non-trivial central extension by Z/2. In this latter case, show that this must be the universal central extension of G. (6) Let G be the inverse image of G in SU(2). This is a central extension of G by Z/2, called the binary icosahedral group. It is a group of order 120. Show that G -+ G is not trivial, by using (2) of this Exercise and (2) of Exercise 4.1.26. (7) Deduce from (4) and (5) that G is the universal central extension of G, that H2(G, Z) g Z/2, and that G is perfect. (For another proof that G is the universal central extension of G, you can show that Hs(G, Z) = 0 using (3) of Exercise 4.1.26.) Since SU(2) can be topologically identified with S3, it follows that SU(2)/G is a compact S-manifold such that ~1 (SU(2)/G) is perfect, hence with Hi(SU(2)/G) = 0. It follows from Poincare duality that H#W(B)/G) = 0 as well. Thus SU(2)/G has the same homology groups as S3, and is known as the Poincar& homology 3-sphere. 4.1.28. Exercise. This exercise will provide some more finite examples of universal central extensions. (1) Show from the identity

(; d!!l) (:, ;“) (d;’ ;) = (:, (d2;ly and its transpose that if F is a field with more than 3 elements (so that there is an element d E FX with d2 # l), then SL(2, F) = E(2, F) is a perfect group. (2) Show that if F, is a finite field with Q elements, then SL(2, F,) has order q(q2 - 1). Note that in fact the restriction in (1) on the cardinality of F is necessary, since SL(2, F2) 2 Ss and SL(2, F’s) is a solvable group of order 24. (3) From (1) and (a), SL(2, F4 ) is a perfect group of order 4(15) = 60. Show that it is isomorphic to the group G g A5 of Exercise 4.1.27 by showing that G acts faithfully as a permutation group of the set of 5 elements i?l(F4) =&f (F; \ ((0, o)})/F,x g IF4 u {m}.

Deduce from Exercise 4.1.27 that Hs(SL(2, lF’4), Z) has order 2, so that its universal central extension is an extension by Z/2.

1. Universal central extensions and H2

185

(4) From (1) and (2), SL(2, F5 ) is a perfect group of order 5(24) = 120 = 3 * 5 . 23. Clearly the Sylow S-subgroups and Sylow 5subgroups of this group are cyclic. Show that the Sylow 2-subgroups are isomorphic to Q, which has vanishing Hz by (3) of Exercise 4.1.26. Deduce that Hz(SL(2, IFS), Z) = 0 and that G is its own universal central extension. In fact SL(2, Fs) is isomorphic to the binary icosahedral group G of Exercise 4.1.27. ( 5 ) GL(3, IFS) = SL(3, F2 ) is a perfect group of order (23 - 1)(23 2)(23 - 4) = 7 .6 .4 = 3 .7 .8. Clearly the Sylow 3-subgroups and Sylow ‘I-subgroups of this group are cyclic. Show that the group of upper-triangular matrices with l’s on the diagonal is a Sylow 2subgroup isomorphic to a dihedral group of order 8. Deduce that WSL(3, p2), 4 is a finite abelian 2-group, and see if you can compute it. 4.1.29. Exercise. Show that group homology commutes with direct limits (cf. Theorem 1.2.5). In other words, if (Ga)crE~, (e,, : G, + Gp)a

1 and Gi is an abelian group written additively. Show that

Deduce using the structure theorem for finitely generated abelian groups and induction on the number of finite cyclic summands that there is a natural isomorphism i&(G,Z)“? \G for any finitely generated abelian group. Then use Exercise 4.1.29 to conclude that this is valid for any abelian group. Note that this calculation is consistent with the calculation that Hz(I.7, Z) 2 Z/2 in Exercise 4.1.26(l).

2. The Steinberg group We’re now ready to apply the theory of the previous section to the perfect group E(R) of matrices over a ring R. Recall from Lemma 2.1.2 that this has generators eij (a), i # j, a E R, satisfying the relations eij(a)ejj(b) = ej.j(u +

(4

b);

eij(u)%(b) = w(b)eij(a),

j#

k and i # 1;

eij(u)ejk(b)eij(u)-‘e~~(~)-’ = ea(ub), i,j, k distinct; eij(u)eki(b)eij(u)-‘ear-’

= ekj(-bu),

i,j, k distinct.

(b) (c) (d)

4.2.1. Definition. Let R be a ring. For n 2 3, we define St(n, R), the Steinberg group of order n over R, to be the free group on generators xii(u), i # j, 1 I i, j I n, a E R, divided by the relations Xij(U)xij(b) = xij(U + b);

(4

k and i # 1;

(b)

xij(u)x~~(b) = xkl(b)xij(u),

j #

~i~(U)~~~(b)x~~(u)-lx~~(~)-’

= xik(~b), i,j, k distinct;

X~~(U)X~~(~)X~~(U)-'X~~(~)-~

= xkj(-bu),

i,j,

k distinct.

(c)

(d)

It is immediate that St(n, R) is a perfect group and that there is a unique surjective homomorphism (Pi : St(n, R) + E(n, R) satisfying xij(u) H

188 4. Milnor’ s Kz

eij(a). Clearly there are natural maps St(n, R) + St(n + 1, R). However, unlike the situation with the maps GL(n, R) + GL(n + 1, R), it is not clear that these are injective (and in fact this is not always the case). We let St(R), called simply the Steinberg group of R, be the inductive limit. In other words, this is the universal group on generators xij(a), i # j, 1 5 i, j < co, a E R, satisfying the above relations. By construction, there is a canonical map cp : St(R) ++ E(R) (the limit of the (Pi as n -+ CO). The definition can be simplified a bit because relation (d) is redundant. By relation (a), x~(a)-’ = xij(-a). Multiplying (c) on the left by xij(a)-’ and on the right by xij (a), we obtain xj#)x&)-lxj&)-l

Xij(u)=Xij(U)-lXik(Ub)Xij(U)=Xi~(ub)

(using (b) and the fact thata, . j, k were assumed distinct), which is the same as (d) if we renumber indices and replace b by a, a by -b. Thus it suffices to assume (a), (b), and (c). Note also that the group St(R) is functorial in R, since if (L: : R + S is a homomorphism of rings, there is a unique map from the free group on generators zij(u), a E R, to St(S) sending zij(u) to xij(a(a)), and since this is compatible with the relations in St(R), it factors through a map (Y* : St(R) + St(S). 4.2.2. Definition. Let R be a ring. We let KS(R) = ker(cp : St(R) -++ E(R)). This is functorial in R since the groups E(R) and St(R) and the homomorphism ‘p are functorial. The rationale for this definition is that Kz(R) vanishes precisely when all relations among matrices in E(R) follow from the “obvious” relations of Definition 4.2.1. Thus Kz(R) measures the “non-obvious” relations among elementary matrices over R, just as Kl(R) measures the failure of general invertible matrices to be expressible in terms of elementary matrices. 4.2.3. Lemma. Let R be a ring and 3 5 n < 00. The subgroup of St(n, R) generated by all Xij(u), u E R, with i < j is nilpotent, and (Pi restricted to this subgroup is an isomorphism onto the upper-triangular subgroup of E(n, R). Thus K2(R) has t rivial intersection with the sub group of St(n, R) generated by all zij(e), a E R, with i < j. Proof. Let N(n, R) be the subgroup of St(n, R) generated by all xij(a), a E R, with i < j. This contains the subgroup Nr generated by all xrj(a), a E R, 1 < j 2 n. By relations (a) and (b), Nr is abelian and R”- ’ surjects onto Nr via (u2, u3, "', u,) - X12(u2)X13(u3)~'~xl~(urz)~

But under (Pi, Nr maps to the upper-triangular matrices with l’s on the diagonal whose other non-zero entries are all in the first row, so the composition RnP1 --H Nl % E(n, R) is injective and (Pi must be injective on Nr . By relation (c), N(n, R) normalizes Nr . Let N2 be the subgroup of St(n, R) generated by all xij(e), a E R, with i < j and i = 1 or 2. Then

2. The Steinberg group

189

Nz/Ni is generated by the images of the xzj(a), a E R, 2 < j 5 n. Arguing as before, the group generated by these is also abelian and an image of Rnw2, and maps to the upper-triangular matrices with l’s on the diagonal whose other non-zero entries are all in the second row. So (Pi is injective on this group as well and so on Nz. Continuing inductively, one sees that N(n, R) is an iterated extension of abelian groups and maps isomorphitally under (Pi to the group of upper-triangular n x r~ matrices with l’s on the diagonal. 0 4.2.4. Theorem. Let R be a ring. Then K2(R) = ker(cp : St(R) -t E(R)) is an abelian group, and is precisely the center Z(St(R)) of St(R). Thus St(R) is a central extension of E(R). Proof. Let x E Z(St(R)). Then ‘p( z ) must commute with p(y) for all y E St(R), and since cp is surjective, v(x) E Z(E(R)). But E(R) has trivial center, since an n x n matrix can’t commute with each eij(l) unless it is a diagonal matrix and all its diagonal entries are equal, and E(R) consists of infinite matrices whose diagonal entries are eventually 1. So cp(z) = 1, showing that Z(St(R)) C Kz(R). For the reverse inclusion, suppose z E Kz(R), and write

2 = Xi& (ai) . . . xi,j,, (4, where ei,j, (al) . . . ei,,j, (4 = 1 in E(R). Choose N larger than all the indices ii,. . . , i,, j,, . . . , j,. Then for any 1 5 n, any lc < N, and any a E R, xiljl (~l)xkN(++j,

(al)-l =

xkN(a),

k # .h,

xi,N(ala>zkN(a>,

k =jl,

so x normalizes the subgroup AN generated by the zkN(a), k < N and a E R. But by Lemma 4.2.3, the restriction of cp to AN is injective. So for Y E AN, h?/-lY-l) = (P6+&h-'(~c>-'(Pb)-1 = cp(Y)cP(Y)-’ = I and zyz -’Y -’ = 1. This shows x commutes with zkN(a) for any N larger than all the indices ii,. . . , i, and ji, . . . , j,, for any k < N, and for any a E R. Since these generate St(R) because of relation 4.2.1(c), x is central. Thus Kz(R> C .Wt(R)).

0

4.2.5. Example. Let R be any ring, and let z =

(x12(1)x21(-1>~12(1>)4.

Then

.(x)=((i :)(:1 Y)(i :))’ =pl ;)“=(; :)1 so x E K2(R) and hence x is central in St(R). It will turn out that when R = Z, x has order 2 and generates Ks(Z).

We are almost ready for the main theorem of this section, but first we prove a number of easy group-theoretic identities.

190 4. Milnor ’s K2

4.2.6. Lemma. Let G be a group and let u, v, w E G. Denote the commutator uvu- l21-l by [u, v]. Then (a) [u, v] = [v, u]-‘. (b) [u, vlb, 4 = b, 4[v, [w, 41. (c) (Jacobi identity) If G’ = [G, G] is commutative, then

b, b, 4lb, bA 4l~w, b-4 VII = 1. Proof. (a) is trivial. For (b), note that

[u, vw][v, [w, u]] = u(vw)u-l(vw)-lv(wuw-lu-l)v-l(uwu-lw-l) = (uv)(wu-lw-lv-l)(vwuw-~)u-lv-yuwu-lw-l) = k4 vlb, WI. For (c), first rewrite (b) as

[v, [w, 41 = [UT vwl-v4 Vl[% WI. Cyclically permuting U, v, w and multiplying gives (provided commutators commute)

4.2.7. Theorem. Let R be a ring. Then St(R) is the universal central extension of E(R).

Proof. By Theorem 4.2.2, St(R) is a central extension of E(R), and relation 4.2.1(c) shows St(R) is a perfect group. By Theorem 4.1.3, it suffices to show that every central extension of St(R) is trivial. Let (U, IJ) be a central extension of St(R). If 2, y E St(R) and we choose X, Y E U with $(X) = 2 and q(Y) = y, then [X, Y] is independent of the choices of X and Y, since changing X or Y by an element of Z(U) will not affect the commutator. Thus it makes sense to refer to [+-l(z), $-l(y)] as a well-defined element of U. We will define a splitting map s : St(R) + U to $ by sending [z, Yl H M-74, V(Y)1 for suitable 2 and y.

2. The Steinberg group 191

42.8. Lemma. In this situation, if j # k, i # 1, and a, b E R, then [+-l(zij(a)), $-‘( 324(b))] = 1 in U. Proof. Choose h distinct from i, j, k, 1 and choose

u E tie’(m(l)),

2,

E

$-‘(nj(a)),

and w E ti-‘(w(b)).

Then [u, w] E +-l(zij(a)). There must be elements c, c’ E Z(U) such that cuw = wu, c’vw = WV. Then [[U, 211, w] = (uvu-~Y-l)w(v~w-~~-~)w-~ = ( UVu-‘)c’w(uV-~)(u-rw-l> = c’uwu-1w(u21-1)(w-1U-1c-1) = c’c-luV(nu)(w-lV-l(c’)-l)U-’ = (Uww)(w-l’u-lu-l) = 1. 0 4.2.9. Lemma. In this situation, if h, i, j, k are distinct and a, b E R, then

[tiel(xhj(a)), tie’(xjk@))l = [$~-~(xhi(l)), +-‘(xik(ab))l in U. Proof. By relations 4.2.1,

hj(a>, x&)1 = [m(l), w(ab)l = xdab)

in St(R).

Choose u E +-‘(zhi(l)), w E $-‘(+(a)), and w E $I-‘(zjk(b)). Then [u, w] E $-‘(zhj(a)), [v, w] E $-‘(zirc(ab)), and [u, w] = 1 by Lemma 4.2.8. Furthermore, [u, V] commutes with u, with V, and with [TJ, w] by Lemma 4.2.8. So if G is the group generated by u, V, w, [G, G] is commutative. Now apply (c) of Lemma 4.2.6. We obtain

b, I? wll[c [w, 41k4 b, 41 = 1, or since [w, U] = 1, [[u, v], w] = [u, [v, w]], which is what we want.

•I

Proof of Theorem 4.2.7 (continued). Recall that we want to define a splitting map s : St(R) + U. Since St(R) is given by generators and relations, it will be enough to define elements sij(a) E U, i # j and a E R, satisfying the same relations as the xij(a) E St(R). Then there will be a unique homomorphism s : U + St(R) sending xij (a) I+ sij(a), and pro vided we choose sij(a) E +-‘( xij(a)), s will split $J and thus demonstrate that (U, $J) is a trivial extension. So let a E R, i # j. Choose k distinct from i and j and define

192 4. Milnor’s Kz

By Lemma 4.2.9, this is an element of $-‘(x~(a)) independent of the choice of k. We will show that the elements Sij(a) satisfy the relations 4.2.1. Lemma 4.2.8 immediately gives relation 4.2.1(b). To check 4.2.1(a), let a, b E R and choose lc distinct from i and j. Choose u E +-‘(qk(l)), u E +-‘(~/cj(a)), w E $-l(ICkj(b)). Then by Lemma 4.2.6(b),

sij (a)sij (b) = [F ~1

[u, W] = [u, VW] [u, [w, u]].

But [w, U] commutes with u by Lemma 4.2.8 and

VW

E $‘-l(xkj(a + b)), so

sij(a)sij(b) = [‘% VW] = [ti-‘(Gk(l)), $-‘(zkj(a + b))], which by definition is sij (a + b). Finally, we need to check relation 4.2.1(c), but this follows immediately from Lemma 4.2.9. 0 4.2.10.

Corollary. If R is any ring, there is a natural isomorphism

Kz(R) --+ Hz(E(R), Z).

Proof. This follows immediately from Theorems 4.1.19 and 4.2.7.

0

4.211. Remark. Something that comes out of the construction used in the proof of Theorem 4.2.7 is that if 2 and y are commuting elements of E(R), then [V’(X), F’( Y)] is a well-defined element of St(R) which maps to [z, y] = 1 under cp, in other words, an element of Kz(R). In fact, this is the most useful way of constructing elements in Kz(R), and under favorable circumstances, K2(R) is generated by such elements. A case of particular interest is when R is a commutative ring. Then the units of R, RX, form an abelian group, and for u E RX, : ,yl E ( > E(2, R) by Corollary 2.1.3. 4.2.12. Definition. Let R be a commutative ring, and let u, 21 E RX. The Steinberg symbol {u, v} is defined to be the element [(p-‘(d12(~)), ~-‘b-h(~))1 of WR) ( as in Remark 4.2.11), where

(Since dlz(u) and dls(w) commute in E(R), this indeed defines an element of Kz(R).) Note from the identities used in Corollary 2.1.3 that d12(u) = e12(21)e21(--21-1)e12(u)e12(-1)e21(1)e12(-1), d13(2)) = e13(w)e31(-21-1)e13(21)e13(-1)e31(l)e13(-l), SO

that if we define wij(u) E St(R) and hij(u) E St(R) by Wij(u) =def zij(U)zji(--‘zL-1)2ij(‘LL),

hij(u) =&f

Wij(th)Wij(-I),

2. The Steinberg group 193 then

and {% ?J}

=def [hl2(u), h3(V)].

The Steinberg symbols can also be described in terms of group homology. Note that if G is a free abelian group on two generators s and t, then ZG FZ Z[t, t-l, s, s-l], and the trivial G-module Z has the free resolution

Furthermore, if 0 is the automorphism of G interchanging t and s, then the following diagram commutes: (‘-l>

0 - ZG

I

-0

0

ZG ~ ,&,

,

u (s-Ll--t),

,

(t-1ts-1)+

ZG

t-1, SC1

ZG

t-1

.-&

u 0

( >l

1

0 - ZG

l-‘),

ZG ~

ZG

(t--l,

,

e-+1

?&

Thus H.(G, Z) is the homology of the complex

with IS acting by -1 on the first Z and interchanging the two summands in the Z2, and in particular Hs(G, Z) is free abelian on a generator that is sent to its inverse if we interchange t and s. Thus Hz(G, Z) is naturally isomorphic to A2 G, the alternating tensor product or second exterior power (this is a special case of the result of Exercise 4.1.31). The commuting elements di2(~) and d13(v) of E(R) define a map cx : G -+ E(R) with t I-+ d12(u), s H d13(w), and cy* sends the canonical generator t A s of Hz(G, Z) to the Steinberg symbol {u, V} in Hs(E(R), Z) Z Ks(R). In fact, the diagonal matrices in E(3, R) are an abelian subgroup isomorphic to (RX)2 (the determinant must be 1, so the (3, 3)-entry is determined by the (1, 1) and (2, 2)-entries), generated by elements of the form d12(u) and d13(v). By Exercise 4.1.32, Hz((R~)~, Z) g A2((Rx)‘), which evidently contains (RX x (1)) A ((1) x RX) as a direct summand. The subgroup of Ks(R) generated by the Steinberg symbols is the image of (RX x { 1)) A ({ 1) x RX) in Hz(E(R), Z) % Kz(R). 4.2.13. Example. Of course, the whole definition would be a little silly if {u, 7~) were always trivial. However, if u = 21 = -1 and R = Iw, then d12(u) and d13(w) generate a Klein 4-group in SO(3) c SL(3, Iw) (see Exercise 4.1.26), and the inverse image of this 4-group in the universal covering group SU(2) c SLZ) is a quaternion group Q. The same holds if we first embed SL(3, R) in SL(n, IF%) for any n > 3 and then take

194 4. Milnor ’s K-J

the universal cover SLTR), since the embedding SO(3) L) SL(n, IF!) induces an isomorphism on ~1 and thus the induced mapping SU(2) + SL?R) is injective (compare Example 1.6.13 and Examples 4.1.5). Since 1% SLrl) is a quotient group of St(R), this proves that { -1, - 1) maps to an element of order 2 in the corresponding quotient of Ks (Iw). In fact this quotient group splits (see (2) of Exercise 4.1.26, or else note that by Lemma 4.2.14 below, (-1, -1) can have order at most 2), so (-1, -1) E Ks(Iw) has order exactly 2. 4.2.14. Lemma. Let R be a commutative ring. The Steinberg symbol mapRXxRX a Kz( R) is skew-symmetric and bilinear, that is, {u, w} = {v, ~1~’ and {WW, v) = (~1, u}{.zL~, v}. Proof. This is immediate from the homology approach, since as mentioned above and proved in Exercise 4.1.32, H2(G, Z) 2 A2(G) for G an abelian group. Alternatively, we can check this directly from the definition above, since 1 0 0 (p(w23(1))

=

0

0

1

.

0) ( 0 -1 conjugates dis(zl) to die and vice versa. To prove skewThus (P(w23(1)) symmetry, note that {v, u}

=def [(p-1(dl2(7’) ), =

+@13(‘L L))]

[W23(1)(P-‘( h3(~))W23(1)-1, W23(1)(P-‘( h2(~))W23(1)-1]

= W23(1)[(P-‘( h3(v)), +(d12(U))]W23(1)-1 = W23(1){% w}-’ W23(l)-l = {U, 21)-l.

Here we have used Lemma 4.2.6(a) and the fact that KS(R) is central in St(R). To prove bilinearity, note that by Lemma 4.2.6(b), {‘(I,

01212) =def [(P-1(h2(u)), +(&3(Vlw2))] = [+(&2(u)), ‘P -1(d13(~l))+(d13(~2))] =

[+(&2(U)), ‘f -1(d13(~1))][(P-1(~12(u)),

+@13(w2))]

[(p-1(dl3(%)), [+(&3(V2)), (o-‘( h2(u))]]-’ =

{u, wl}{% V2}[(P-1(d13(~l)), 1% w2}-1]-1

= (‘1 1, VI){% v2),

again since Ks(R) is central. Bilinearity in the other variable follows from the skew-symmetry. 0 Most of the rest of this section will be taken up with calculations using the relations (4.2.1), in order to give a slightly more convenient description of the Steinberg symbols and in order to prove that they satisfy two additional relations. These relations are important for the applications of K2(R) in the next section.

2. The Steinberg group

195

4.2.15. Lemma. If R is any ring and u E RX, the elements wij(u)

&j(u) =def W&)“ij(-1)

=def zij(‘1L)zji(--u-1)2ij(‘u.),

of St(R) defined in 4.2.12 satisfy (w&))-1 = (m&u)),

Wij(U) = uJj+u-l),

hij(l) = 1. In addition, if u, v E RX and lc # 1, i # j, then w&J), ~k1(4%(4 (wl(4)-1 =

i, j, k, 1 all distinct,

2uu (-ZAJ), k = i, i, j, 1 all distinct, w,l(_vu) 2 7 k = j, i, j, k all distinct, w~~(-u-~vu-~), k = i, j = 1.

Proof. To begin with, by (4.2.1)(a), %j(“)Wij(-21) = zi~(u)~cji(--u-~)z~j(‘u)z~j(--21)zji(u-~)2ij(-~) = zi~(u)“~~(--‘u-‘)zj~(21-~)zij(--u) = Zij(U)Z&U) = 1, so (wjj(u))-l = wij(-u). In particular, hi = zuij(l)wij(-1) = 1. The fact that wkl(u) and wij(v) commute if i, j, k, 1 are all distinct is obvious from (4.2.1)(b). Next suppose i, j, 2 are all distinct and k = i. We have by (b), (c), and (d) of (4.2.1) WiZ(.u)G&) (%l(u))-l = ~i1(2C)21i(--U-l)5iz(‘LL) (Zij(V)) ~cil(-~)21i(~-‘)~il(-u) = z&)2&?r1) (zij(v))xli(u-‘)~il(-2l) = 2il(U) (2~j(-u-1zJ)~&J)) Q(-u) = xij(-~)21j(--21-1’u)~cij(21) = z&u%). Similarly

196 4. Milnor ’s Kz

so WZ(U)Wj(W)

(wiz(‘zL))-l = w~~(u)Icij(w)Icji(-w-1)2ij(w) (Wil(U))_l = qj ( -u-lw)zj~ (w-lu)zlj (-u-lw) = wu (-u-h)

and

which gives the second and third relations. Finally, to get the last relation, choose 1 distinct from i and j and note that by what we’ve already proved, wij(w) = wil(l)wu (w)w~~(-1). So %(4%(4 (wj(u))-l = w&) (wil(l>~z~(v)2uil(-l)) (w&))-l = w~~(-2L-~)w~~(wu-~)w~~(zL-1) = zuj~(-U-lwu-l). Taking u = v in this relation gives

w&u) =

wj~(-u-‘uu-‘) = w+(-u-l).

q

4.2.16. Corollary. If R is a commutative ring and u, w E RX, then

In other worck, if we identify RX with a subgroup of E(R) via u H dlz(u), then h12 gives a section RX + St,(R), and the Steinberg symbol is the inverse of the associated cocycle in Z2(RX, Kz(R)) as defined in the proof of Theorem 4.1.16. Proof. We have (‘2 1, u) = blZ(U), h3(w)l = h2(+13(~)(h2(u))-1(h13(w))-1 = hl2(u)Wl3(w)W13(-1)

(w12(1)w12(-u)) w13(1)w13(-w)

= h2(U)W13(V) (w32(1)w32(-u)) %3(-w) = hl2(U)Wl2(~)W2(+J~) = hl2(U)W12(w)W12(-1)W12(1)W12(-21’1 1) =

h2(74h12(4(h2(74)-1.

0

2. The Steinberg group

197

4.2.17. Theorem. If R is a commutative ring, the Steinberg symbol map

RX x RX (‘1 Kz (R) satisfies the additional relations (a) {q--U}=lforUERX, (b) {u,l-u}=lforu~R~,l-UER~. Proof. (a) By Corollary 4.2.16, we need to show that h12(~)h12(-~) = h12(-u2). But by the last identities of Lemma 4.2.15, h12(U)h12(-U)

= w12(~)~12(-l)~12(-~)w2(-1) = w21(~-“) w2(-1) =

W~2(-K2)W12(-1) =

h12(-4.

(b) By Corollary 4.2.16, we need to show that h12(u)h12(1- tL) = h2(u - u”) .

But hz(u)h2(1 - u) =w12(u)w12(-1)~12(1 =wl2(~)w21(1)w12(1

- u)w2(-1) - u)w2(-1)

=w12(u)221(1)~12(--1)~21(1)w12(1

- uh2(-1)

= (W12(2l)Z21(1)W12(-~)) W2(~)Q2(-1)W12(1 - u) (w12(u - 1)221(1)w2(1 - 7J)) %2(-l) ‘~ 12(-~2)w~2(~)~12(-1)2u12(1-

u)212(-(1 - 42)w12(-1)

=212(-u2)212(4521(-7J-1)3a2(4

x12(-1)21112(1 - 21)212(-(l- u)2)w12(-l) =212(u - u2)z21(-U-1)z12(u =212(u - u2)521(-K1)512(u

x21(-(1 -

l)w2(1- u)a2(-(1 - 42)w(-1)

- 1)x12(1 - u)

u)-1)212(1 - u)z12(-(1 - 42)w12(-1)

=212(u - u2)~21(-u-1)221(-(1 - 4-7512(21. - .112)w2(-1) =212(u - u2)221(-u-1(1 - 2L)-7212(2L

=w12(u(l- U))W12(-1) = h12(u - u’).

- u2)w12(4)

q

4.2.18. Corollary. If R is a finite field, all Steinberg symbols vanish in

K2(R). Proof. Let R = IF,, the finite field of q elements. Since JF: is cyclic of order q - 1, we may choose a generator u for IF:, and by bilinearity of the

198 4. Milnor’ s Kz

symbol map, it suffices to prove that {u, U} = 1 (we are using multiplicative notation for Kz). By skew-symmetry of the Steinberg symbol, {u, U} = {u, u}-l, i.e., {u, u} has order at most 2. If q is a power of 2, then -1 = 1 in IF,, so by (a) of the Theorem, {u, U} = {u, -u} = 1. If q is odd, then by bilinearity and (a) of the Lemma,

(24,

u} = (21, -u}{u,

-1) = {U, -1) = {U, 73) = {U, U}?

So if 9 is even, we again conclude that {u, U} = 1. If 9 is odd, then -1 is not a perfect square in IF,. Suppose we can choose w E IF: such that neither w nor 1 - w is a perfect square in IF,. By (b) of the Theorem, {w, 1 - w} = 1. Butsince neither w nor 1 - w is a perfect square, they are both odd powers of U, so {w, 1 - w} is an odd power of {u, U} and {u, U} = 1. So it’s enough to show a suitable w exists. Since -1 is not a perfect square in F,, we need to show there is a w, not a perfect square, such that w - 1 is a perfect square. But such a w exists, since otherwise adding 1 to a perfect square would always give a perfect square in IF,, and 1, 2, . . .) -1 would all be perfect squares, a contradiction. Cl 4.2.19. Example. If R = Z, then RX has only two elements, 1 and -1. We saw in Definition 4.2.12 that (-1, -1) has order 2, and this is the only non-trivial Steinberg symbol, since {u, w} = 1 if u = 1 or w = 1. In this particular case, relation (b) of Theorem 4.2.17 is vacuous and relation (a) is trivial. As mentioned in Example 4.2.5, the element 2 =

(2~2(1)Ic&l>~c12(l))4 = (w12(l))4

of St(Z) also lies in Ks(Z). However hiZ((_l). (-1))(hr2(-1))-1(hi2(-1))-1

= (-1, -11-l = (-1, -1)

by Corollary 4.2.16, and since wis(l) = (wr~(-l))-~ and his(l) = 1 by Lemma 4.2.15, the left-hand side simplifies to (h12(-1))-2 = (wi2(-1)w12(-1))-2 = (Wi2(-1))-4 = (w&l))4 = 2. So the element of Ks(Z) constructed in Example 4.2.5 is the same as (-1, -1). For more complicated rings, the relations in Theorem 4.2.17 are more interesting. For instance, if R = I&[> Xij(b>(Xiljl(Ul)Xiljl(U2). ' *Xi,j,(G))-'

7

b E I. As we saw above, we are free to replace xij(b) by xij(O) = 1, which then shows y can be made trivial after modification by an element of Ks(R). So ker(8) C im(q,). It remains to check exactness at Ki (R, I). The composite

is 1, since if x E Kz(R/I) and we choose y E Ks(R) with x = q*(y) as above, then cp~(y) E E(R) and maps to 1 in Ki(R) = GL(R)/E(R). Conversely, if g E GL(R, 1) and the image [g] of g in Ki(R, I) maps to 1 in Kr(R), this means g E E(R). So g = cp~(y) for some y E St(R). If x = q*(y) E St(R/I), then (PR/I(x) = q*(g), which is 1 since g E GL(R, I). So x E ker(cpR,I) = Kz(R/I), and a(x) = [g] by construction. Cl So far, we have not been able to compute K2 in very many examples, though at least we’ve produced examples of rings where it is or is not trivial (Example 4.2.13 and Exercise 4.2.24). Our aim next is to study Ks in the case of a (commutative) field. Unlike KI and KO which are not particularly interesting for fields, this is a decidedly non-trivial subject with a lot of applications. However, following ideas in [Keune], Theorem 4.3.1 now gives a way to relate the calculation of Ks of a field to a problem about K1, which can be studied using the theory of relative Mennicke symbols from Theorem 2.5.12.

202 4. Milnor’ s Kz

4.3.2.

Lemma. It F is a field, there is a natural epimorphism d :

K2(F) + SKl(F[tl, (t2 -t)).

Proof. Let R = F[t], which is a PID, and let I = (t2 - t) s R. T h e n R/I = F[t]/(t2 - t) ” F x F, with the quotient map q : R + F x F corresponding to evaluation at 0 and at 1. By Corollary 2.3.3, Kl(R) = FX, and the map q* : K1(R) + K1(R/I) = K1(F x F) 2 FX x FX is obviously the diagonal map, which is injective. Furthermore, since the map R + F corresponding to evaluation at 0 is split surjective via the inclusion of constant polynomials, we get a splitting K2(R) g Kz(F) x NKz(F) as in Theorem 3.2.22. The map q* : K2(R) --) K2(R/I) = K2(F x F) g K2(F) x K2(F) is obviously the diagonal injection on the K2(F) factor, so the cokernel is a quotient of Kz(F). Then d gives an isomorphism of this quotient of KS(F) onto K1 (R, I) = SK1 (R, I) (since any unit in F[t] is actually a unit in F, and thus can’t be 3 1 mod I unless it is equal to 1). 0 In fact, one can show that Kz(R) % K2(F), and d is an isomorphism of KS(F) onto SKl(F[t], (t2 -t)). This makes the calculation of K2(F) essentially equivalent to the calculation of the relations among the relative Mennicke symbols from Theorem 2.5.12. (See Proposition 4.4.2 below.) The key to getting more information is the following theorem, which can be proved either using calculations in the Steinberg group (for a proof along these lines, see [Milnor, $91) or else using homology, as in [Hutchinson], on which the following proof is based. 4.3.3. Theorem. If F is a field, then Kz(F) is generated by Steinberg symbols. Proof. Recall that in the case of a field, E(n, F) = SL(n, F) (Proposition 2.2.2). By Definition 4.2.12 and Exercise 4.2.21, the subgroup of KS(F) generated by the Steinberg symbols is precisely the image of the corestriction map H2(FX x FX, Z) + H2(SL(F), Z), where FX x FX q SL(3, F) -+ SL(F) via (a, b) H (i i a_!b_i). W e willshow t h a t except in the case of a few finite fields of small cardinality, the corestriction maps I&(FX x FX, Z) ---) &(SW, FL z) + I&(%(4, F), z) --f ... are all surjective; in fact Hz(SL(n, F), Z) -+ H2(SL(n + 1, F), Z) is an isomorphism for n 2 3. Since Kz(F) = 15II2(SL(n, F), Z) (Exercise 4.1.29), this will prove the theorem and a bit more. We split the proof into

3. Milnor’s K2 203 several steps; the Theorem is obtained by combining Propositions 4.3.6 and 4.3.11 and Theorem 4.3.12. The first Lemma involves some of the same ideas as Theorem 4.1.20. It (and similar results) is actually most easily proved using the theory of spectral sequences, but we give a direct proof, at least for the case we need. Cl 4.3.4. Lemma. Let G = N K H be the semidirect product of a normal subgroup N by a group H, and let M be a G-module. If H,(H, H,(N, M)) = 0 for all p and q with p + q = j, then Hj(G, M) = 0. Proof.

This is clear if j = 0, since

H,,(H, Ho(N, M)) = (M/{n, - 1: 72 E N})/{h, - 1 : h E H} = M/{g, - 1 : g E G} = Ho(G, M). The general case is reduced to this case by induction on j, using resolutions. For example, we do the cases j = 1 and j = 2, which we will need below. Start by choosing an exact sequence of G-modules 0 + Ml -+ F0 + M -+ 0, with Fo free, and note that Fo is free not just as a G-module but also as an N-module. The corresponding exact sequences in N-homology and in G-homology give Hj+l(N, M) = Hj(N, MI),

Hj+l(G, M) g Hj(G, Ml),

j 2 1,

as well as the exact sequences 0 + Hr(N, M) --) Ho(N, MI) * Ho(N, Fo) + Ho(N, M) --f 0, 0 -+ H1(G, M) + H,,(G, MI ) -+ Ho(G, Fo) + Ho(G, M) -+ 0. Split the first of these into two exact sequences 0 -+ H1(N, M) + Ho(N, MI) --+ K + 0, 0 + K -+ Ho(N, Fo) + Ho(N, M) + 0. Applying H-homology and assuming that Ho(H, HI (N, M)) = 0 and H1(H, Ho(N, M)) = 0, we see that Ho(G, MI) = Ho(H, Ho(N, MI)) g Ho(H, K) and that there is a short exact sequence 0 + Ho(H, K) --+ [Ho(G, Fo) g Ho(H, Ho(N,

+ [Ho(G,

Fo))l

M) g Ho(H, Ho(N, M))] -+ 0.

Comparing this with the exact sequence in G-homology, we see that HI (G, M) = 0. Also, replacing M by Ml lowers j by 1 and enables us to repeat the same trick, thus proving the Lemma by induction. 0

204 4. Milnor’s K2

4.3.5. Lemma. If F is a field, T g FX x FX is the group of diagonal matrices in GL(2, F), and B g T K F is the group of upper-triangular matrices in GL(2, F), then if F is infinite, the corestriction map

coming from the inclusion of the diagonal matrices is an isomorphism. For finite F, this map is still an isomorphism in degrees 1 and 2 if F has more than 2 elements. Proof. Note that B 2 FX x Aff(F), where the first factor corresponds to the scalar matrices, and the second factor, the affine group or “az + b E t . The group B ( ) acts transitively on the set F by letting the scalar matrices act trivially and letting Aff(F) act by afhne transformations. The stabilizer of the point 0 for this action is just the subgroup T. Consider the short exact sequence of B-modules group” of F, is the group of matrices of the form

where ZF denotes the free abelian group on the set F, with B-action coming from the B-action on the set F, cx sends each point of F to 1 E Z, and M = kercu. Since B acts transitively on F with T as one of the stability groups, ZF 2 ZB 63.~~ Z as a B-module, and by Corollary 4.1.12, H.(B, ZF) ” H.(T, Z). By the remarks in Definition 4.1.21, the map (Y* can be identified with the corestriction map in the Lemma. So the Lemma will follow from the exact sequence of Proposition 4.1.9 if we can show that H.(B, M) = 0. For this we apply Lemma 4.3.4, so we need to show H.(T, H. (F, M)) = 0. Since H,(F, ZF) = 0 for q > 0 a n d Ho(F, ZF) + Ho(F, Z) is an isomorphism, H,(F, M) 2 H,+l(F, Z). For instance, Ho(F, M) g F (and this isomorphism respects the T-module structure). If F has more than 2 elements, then there is some a # 1 in Fx , and a - 1 is invertible in F. So F/(a - l)F = 0 and Ho(T, Ho(F, M)) = 0. Similarly, since T is abelian, E y must act by the identity on ( > all homology groups H,(T, Ho(F, M)), whereas a - 1 is invertible, and thus the homology groups are all 0. When F is infinite, the fact that B,(T, B,(F, M)) = 0 f or all q can be derived from this; for instance, if F has characteristic 0, F is torsion-free as an abelian group, and thus H,(F, M) ?E H,+l(F, Z) 2 A”+l(F) by Exercises 4.1.29 and 4.1.31, and a similar argument applies, since a E FX c Aff (F) acts on /\‘+l (F) by aq+’ but must act on homology by the identity. As pointed out by Suslin [SuslinLNM], a slightly different argument is required in the case F is of positive characteristic p. In this case F is a vector space over lFp and it’s enough to show that H,(T, H,(F, IF,)) = 0 for all q > 0. In this case, it turns out that H,(F, lFp) 2 S(F) @ l\(F) (F viewed as a vector space over lFr,, the generators of the exterior algebra having degree 1 and those of the symmetric algebra having degree 2) if p # 2 and H,(F, F,) ZE!

3. Milnor’s K2 205

S(F) (with generators in degree 1) if p = 2. (See also (2) of Exercise 4.1.32.) If F is infinite, one can still prove vanishing of the cohomology by the same sort of argument as before. If F is finite with more than 2 elements, we still have vanishing of Hc(B, M) and of H, (T, He (F, M)) for all s. Since the quotient map B -+ T induces a left inverse to the corestriction map, we only need to show vanishing of He(T, Hi (F, M)) or of Ho(FX, Hz(F, Z)) = Ho(FX, A2(F)) to get an isomorphism through degree 2. Now a E FX acts on A2(F) by multiplication by u2, so vanishing of Ho(FX, Hs(F, Z)) when F has at least 4 elements follows from the fact that there is an element a E FX with a2 - 1 invertible. And when F has 3 elements, Hs(F, Z) = 0 so the vanishing is automatic. 0 4.3.6. Proposition. If F is any field, the corestriction map H2(FX x FX, 25) --+

H2(G-W, F), z>

coming from the inclusion of the diagonal matrices is surjective. Proof. We consider the action of G = GL(2, F) on X = PI(F) = F u {m} by linear fractional transformations. (This may be defined by letting G act linearly on F2 and taking the induced action on P1(F) = (F2 A ((0, 0))) /FX.) Note that G acts transitively on points of X, on ordered pairs of distinct points, and on ordered triples of distinct points. (When F = F2, X has exactly 3 points and G may be identified with the symmetric group of this set.) Let C, be the free abelian group on ordered (n + 1)-tuples (20, . . . , 2,) of distinct points of X, which is a G-module via the G-action on X. Define E : Co ---) Z by sending each x E X to 1 and define d, : C,+I + C, by n+l

dn(xo, . . . ,

2,) =

X(x0, . . . , ?k, . . . , G). k=O

Note that d, o d,+l = 0 and E o dl = 0, so that (C., d) 5 Z -+ 0 is a chain complex. If F is infinite, this augmented complex C. 5 % + 0 is algebraically the same as the augmented ordered simplicial chain complex of an infinite simplex, which is well known to be acyclic. (Or one can easily check this directly, see [Hutchinson, Lemma 11: let z E ker d,_l . Then z is a finite sum of terms (x,“, . . . , xk) and we can choose x distinct from those x!‘s which appear. If y is obtained from z by replacing each (x,“, . . . , zk) by ( 2,x$, . ..) xk), then d,+ly = z.) If F is finite, the complex is still exact at Ce and Ci, and exact at CZ if F has at least 3 elements. (However, the Proposition is true for F = lF2 anyway since in this case G g Ss and H2(G, Z) = 0 by Corollary 4.1.24 and Exercise 4.1.25.) So assume F has at least 3 elements and look at the long exact homology sequences of the short exact sequences (4.3.7)

o+M~‘c ~4,z+o,

206 4. Milnor’ s K2

(4.3.8)

O-MI+C++M~+O,

(4.3.9)

0+M2+C2%Ml+0.

Let B be the upper-triangular subgroup of G, T the diagonal subgroup, and 2 the center (scalar matrices). Since G is triply transitive on X, and B stabilizes 00, T stabilizes (00, 0), and 2 stabilizes (00, 0, l), we may identify the G-modules Cc with ZG@ z~ Z, Cr with ZG @zT Z, and Cz with ZG@zZ. Thus, by Corollary 4.1.12 and the comments in Definition 4.1.21, we may identify H.(G, Co) with H.(B, Z), the map E* with corestriction H.(R z) z H.(G, Z), H.(G, Cl) with H.(T, Z) 2 H.(FX x FX, Z), and H.(G, C2) with H.(Z, Z) 2 H.(FX , Z). Since do : Cr + CO sends (co, 0) H (0) - (co) = w-l. (co) - (co),

w=

we see that H.(G, Cl) E H.(T z) % H.(B, Z) 2 H.(G, Co) is z H cores(w . z - z). Similarly, dr : C’s -+ Cr sends (00, 0, 1) ++ (09 1) - ( ~,~)+(~,~)=~1~~~,~~-~2~~~,~~+~~,~~ for suitable gr, gz E G, so H.(G, is z H cores(g,l .

C2) g W-T Z) -f% z - Q2-’ . z + z),

H.(T, Z) 2 H.(G, C,)

which since 2 is central is just

cores - cores + cores = cores. After making these substitutions, we obtain from (4.3.7) and (4.3.8) the exact sequences (4.3.7’)

... + Hj+l(G, Z) 3 Hj(G, MO) + HJB, Z) D H~(G, z) 3 H+~(G, Mu) +. . .

and (4.3.8’)

. . . + &+l(G MO) -% HJG, MI) + Hj(T, Z) % HJG, MO) -% H&G, MI) + . . . .

By Lemma 4.3.5, the corestriction map Hz(T, Z) x H2(B, Z) is an isomorphism. So we only need to show H2(B, Z) D H2(G, Z) is surjective, which by the exact sequence (4.3.7’) means we need to show that

3. Milnor’s K2 207

a :

fh(G, MO)

-+ H~(B, Z) in (4.3.7’) is injective. The image of this

map is

ker Hr(J3, Z) 5 Hr(G, Z)) = ker (T % Gab = FX) 2 FX. ( From (4.3.7’) and the facts that Hr(B, Z) z Hl(G, Z> is surjective and Ho(B, Z) = Ho(G, Z) is an isomorphism, He(G, Me) = 0. From (4.3.8’), we have the exact sequence

From (4.3.9), we have the exact sequence (4.3.9’) HI(Z, Z) + Hr(G, M) 5 Ho(G, M2) + Ho(Z, Z) z Z 2 &(G, MI) + 0.

Thus Hc(G, MI) is cyclic, and since by (4.3.8’), He(G, MI) + H,,(T, Z) 2 Z is surjective, this latter map must be an isomorphism. Thus iYl(T, Z) % T % H1(G, MO) is sujective. Since the composite p 0 dl, : Hl(Z, Z) ?f

{(;

i):a~F~} -+H1(T,Z)=

:a,bEFX

is the corestriction map, which is inclusion of the scalar matrices, imp contains the scalar matrices, and T % Hl(G, MO) kills the scalar matrices. Since we already concluded that de* is surjective and that imcr = 2’ n SL(2, F), this means LY must be injective. q 4.3.10. Lemma. If F is any field, there is a natural splitting H2(GL(n, F ) , Z) s H2(SL(n, F), Z) @ H2(FX, Z)

for n 2 3. If F has at least 4 elements, there is an analogous fact for n = 2: H2(GL(2, F), Z) 2 fJo(FX, H2(SW, F), Z)) @ H2(FX, z),

and the corestriction map of Proposition 4.3.6 maps onto the first factor in this decomposition. Proof. For any n, GL(n, F) is the semidirect product of the scalar matrices, isomorphic to FX , and of the normal subgroup SL(n, F). So the

208 4. M i l n o r ’ s K2

inclusion of the scalar matrices, together with the determinant map, gives a split copy of Hz (Fx , Z) inside Hs(GL(n, F), Z). Also, we know SL(n, F) is perfect for n > 3, and this also holds for n = 2 if F has at least 4 elements, by Exercise 4.1.28(l). Since the composite SL(n, F) w GL(n, F) 2 FX is trivial, the corestriction map ff2(Wn, FL z) + ff2(G~Yn, F), z)

has its image contained in the complement of the split copy of Hs(FX , Z). It in fact surjects onto this complement, and gives an isomorphism of the complement with Ho(FX, H2(SL(n, F), Z)), by an argument similar to that in Lemma 4.3.4, since H1(FX, H1(SL(n, F), Z)) = 0. (This is where we use the fact that SL(n, F) is perfect.) To conclude the proof, we need to show that FX acts trivially on H2(=(n, F), z>

for n 2 3. The case n = 2 follows from the next Proposition, since it will turn out that the corestriction map H2(GW, FL z) + ff2(SW~, F), z)

induced by A L) form

is surjective. Since matrices of the give another complement to SL(3, F) inside

GL(3, F) which commutes with matrices of the form A E GL(2, F), the conjugation action of FX on Hs has to be trivial. The case of larger n will then follow from the Stability Theorem (Theorem 4.3.12). 0 4.3.11. Proposition. If F is a field with more than 7 elements, then the corestriction map

fb(GW, F), z) --t WSW F), z) A 0 is surjective, and Hs(SL(3, F), Z) is 0 (det A)-l generated by Steinberg &mb&. if F has more than 3 elements, the corestriction map is still surjective except perhaps for p-torsion, p the characteristic of the field.

induced by A L)

Proof. The general idea is similar to that of Proposition 4.3.6. Consider the action of G = SL(3, F) on X = P2 (F), the set of one-dimensional subspaces of F3. Again let C, be the free abelian group on ordered (n + l)tuples (~0, . . . , x,) of distinct points of X, but with the extra condition

3. Milnor’s K2 209

that if n 2 2, no three xj’s are colinear. This is a G-module via the G-action on X. Define E : Co --+ Z and d, : C,+l + C,, getting an augmented chain complex as in the proof of Proposition 4.3.6. If F is infinite, this augmented complex C. -% Z -+ 0 is acyclic by almost the same proof as before. If F is finite, the complex is still exact at 6’0 and Ci. Note that G acts transitively on X, with the stability group at [es] (er, es, ea the usual basis vectors for F3) : A E GL(2, F), Y E F2

,

transitively on ordered pairs of distinct points in X, with the stability group at (k21, b31) a1

PI=

YI I( Y2

0

0

a2 0

0 :~I,~~EF~,YI,Y~EF , kw2r1 1 1

and transitively on ordered generic triples of distinct points in X, with the stability PUP at ([el, [et], [es])

We proceed as in the proof of Proposition 4.3.6, using short exact sequences of the form (4.3.7-4.3.9) and the corresponding long exact sequences in homology. The substitutes for (4.3.7’-4.3.9’) in our context are as follows: (4.3.7 ”)

... + L$+i(G, Z) -% HJG, A&,) ---f Hj(P, Z) = Hj(G, Z)

2 Hj_l(G,

MO) -+ ... ,

and (4.39’)

H1(FX x FX, Z) -+ HI(G, MI) 3 Ho(G, M2) + Ho(FX x FX, Z) Z il-% Hi,(G, A&) + 0.

We also compute do* and dl, as in the proof of Proposition 4.3.6. Since do : Ci + Co sends (k21, k31) ++ k31 - k21 = [e31 - we1 . k31,

w =

( ) 1

0

0

0

0

-1

0 1 0

)

210 4. Milnor’s

Kz

we see that H.(G, C,) = H.(Pl, z) % H.(P, Z) = H.(G, Co) is z H cores(w . z - z). Similarly, dr : CZ + Cl sends &I, [e21, [e31) ++ (k21, k31) - (My k31) + (hL

=

([ez],

MI

[e31) + 91 . ([e21, [e31> - 92 . (k21, k31)

with

H.(G, C2) ” H.(FX x FX, Z) ++ H.(Pl, Z) s H.(G, Cl) is z H cores(g,l . z - 92-l . z + z). The proof is then exactly the same as that of Proposition 4.3.6, with the following exceptions:

(1) (2)

Since G is perfect, Hr(G, Z) = 0. Assuming F has at least 4 elements, SL(2, F) is also perfect, so Hr(P, Z) g FX. One also has Hr(Pr, Z) = FX x FX. The idea of the proof will be to show that &(P, Z) = H2(G, Z) is surjective, by using (4.3.7”) and showing that cx : H1(G, MO) --) H1(P, Z) Z FX is injective (o is automatically surjective since Hr (G, Z) = 0). So one needs to know as well that ff2(H, Z) = H2(P, Z)

is surjective, where : A E GL(2, F)

” GL(2, F).

Since P = H K F2, we need an analogue of Lemma 4.3.5. This is proved with the same technique, the only difference being that the action of GL(2, F) on F2 is by A$ = (det A)At. So a scalar matrix a 0 acts by multiplication by a3 instead of by multiplication (0 a> by a as in the proof of Lemma 4.3.5. This is no problem as long as

3. Milnor ’s Kz 211

(FX ( does not divide 6, in particular, if F has at least 8 elements. For a smaller finite field of characteristic p, it’s still true that

is an isomorphism, since H.(F2, Z[ i]) vanishes except in degree 0 (by Theorem 4.1.23). (3) In the last step of the proof, one has to examine the composite po (&), : H1(FX x FX, z) +Hl(Pl, z). This time, this is not the corestriction map (which is an isomorphism) but rather the map z H cores(g;’ . 2 - Q2-l . z + 2). If we identify the HI groups with groups of diagonal matrices, this becomes the map

On the other hand, the map (do)* : Hl(f’l, Z) -+ Hl(J’, z> becomes the map

(i % (ab;_l) ++ (; 8 (abi-l) (i (a!-’ y’ = (s ai ._%,)7 whose kernel is precisely the image of the previous map. So a is again an isomorphism as in the proof of Proposition 4.3.6. The fact that Hs(SL(3, F), Z) is generated by Steinberg symbols now follows from combining this result with Proposition 4.3.6. 0 4.3.12. Theorem (Stability for K2). If F is a field, the corestriction maps Hz(WW F), z) + H,(SL(n + 1, F), Z) and Hz(GL(n, F), Z) --+ Hz(GA(n+l, F), z) are isomorphisms for n > 3 if F is infinite, and are at least surjective after inverting the characteristic of the field if F is finite.

212 4. Milnor’ s Kz

Hence (for an infinite field) Kz (F) ?z! Hs(SL(3, F), Z) and is generated by Steinberg symbols. Proof. Because of Lemma 4.3.10 and the fact that the split copy of Hz (FX , Z) in Hz (GL(n, F), Z) clearly maps to the corresponding copy in Hz(GL(n + 1, F), Z), while Ho(FX, H2(SL(n, F), Z)) maps to Ho(FX, Hz(Wn + I, F), z)), it is enough to treat the case of GL(n). The proof of surjectivity, or of surjectivity after inverting the characteristic of the field if F is finite, is virtually identical to the proof of Proposition 4.3.11, except that we use the action of GL(n + 1, F) on X = B”(F), the set of one-dimensional subspaces of Fn+‘. (Inverting the characteristic trivially yields the analogue of Lemma 4.3.5, that the corestriction map

H.(GL(n, F),

+I) -+ H.(GL(n, F) K Fn, d])

is an isomorphism, since Fn is a pgroup and thus its homology with coefficients in Z[i] vanishes by Corollary 4.1.24. For an infinite field, the old proof still works.) Note incidentally that surjectivity of the corestriction map Hz(SL(3, F), z) 4 H2 (SL(n + 1, F), Z) implies because of Proposition 4.3.11 that the latter is generated by Steinberg symbols. The proof of injectivity is only slightly more delicate. For this part of the argument, assume F is infinite, let G = GL(n + 1, F), and let Ck be the free abelian group on ordered (Ic + 1)-tuples (~0, . . . , zk) of distinct points of X = pn(F), butwith the extra conditions that if lc 2 2, no three xj’s are colinear, if Ic > 3, no four zj’s are coplanar, etc. This yields an augmented complex C. 5 Z + 0 which is acyclic by almost the same proof as before. As in the proof of Proposition 4.3.11, let P = PO be the stabilizer of [e,+i], PI the stabilizer of ([e,], [e,+i]), Ps the stabilizer of ([en-11, [em], [e,+il), etc. T h e n CO 2 ZG @zp Z, Cl E ZG @zpl Z, C2 E ZG @zq Z, Cs 2 ZG mzp3 Z, Cd % ZG &p, Z. By Corollary 4.1.12 (Shapiro’s Lemma), H. (G, Cj) g H. (Pj, Z) for j 5 4, and by the analogue of Lemma 4.3.5, the corestriction maps H.(GL(n, F) x FX, Z) + H.(P, Z>, H.(GL(n - 1, F) x FX x FX, Z) ---) H.(Pl, Z), H.(GL(n - 2, F) x FX x FX x FX, Z) + H.(P2, Z), etc. are isomorphisms. So H2(P, Z) z Hz(GL(n, F), Z)@(F’ @FX)@A2(FX). We again use the exact sequences (4.3.7”-4.3.9”)) so we need to show the image of H2(G, MO) -+ Hz(P, Z) does not meet the copy of Hz(GL(n, F), Z) in the latter. Now the composite K(P2, Z> + fL(G, MI) 4 H.(Pl, z)

3. Milnor ’s Kz 213

induced by dl is given by z H cores(g,i . z - ggl . z + Z) and the composite H.(Pi, Z) + H.(G, MO) -+ H.(F, z) induced by do is given by z w cores(w . z - z), as in the last proof. A long diagram chase then shows that the map Hi(P2, Z) + Hr(G, Mi) is surjective and that the map He(G, Ms) -+ He(Ps, Z) 2 Z is an isomorphism, hence that the map Hi(Ps, Z) -+ Hi(G, Mz) is surjective. So the kernel of the map Hi(Pz, Z) + Hl(G, MI), which by the exact sequence (4.3.9”) is the image of the map Hl(G, M2) --+ Hl(P2, Z), is also the image of the map Hl(P3, Z) + HI (Pz, Z). A calculation shows that this coincides with the kernel of the map Hl(Pz, Z) -+ Hl(Pl, Z). So the map Hz(G, MO ) -+ Hl(G, Ml) must be zero and Hz(9, Z) -+ Hz(G, MO ) is surjective. Finally, the image of the map Hz(G, MO) -+

H2(p, Z>

in the exact sequence (4.3.7”) is the same as the image of (do), : Hz(Pi, Z) + Hz(P, Z). Prom the description of (do), as coreso(w, - l), this has trivial intersection with the copy of Hs(GL(n, F), Z) in the latter, which proves what we wanted. Cl 4.3.13. Corollary. If F is a finite field (with the possible exception of Pz), then X2(F) = 0. Proof. By Corollary 4.2.18, all Steinberg symbols vanish, yet Kz(F)[$] (where p is the characteristic of F) is generated by Steinberg symbols by Theorem 4.3.3. On the other hand, for any n >_ 3, SL(n, F) is a finite group, so Hz(SL(n, F),Z) is a finite abelian group by Theorem 4.1.23, whose pprimary part comes from the Sylow psubgroup by Corollary 4.1.24. Now if the order of F is q = pr, the order of SL(n, F) is (qn - l)(qn - q). . . (qn - qn-2)qn-l = q1+2+...+(n-l)(qn - 1). . . (q2 - I), so the largest power of p dividing this is q1+2+“‘+(n-1), which is the order of the subgroup N(n, F) of upper-triangular matrices with l’s down the diagonal. Thus N(n, F) is a Sylow p-subgroup of SL(n, F). However, by Lemma 4.2.3, there is a homomorphism N(n, F) --) St(n, F) which splits the canonical map ‘p : St(n, F) + SL(n, F) over N(n, F). This shows that the central extension cp : St(F) -+ SL(F) is trivial over N(F) = l%N(n, F), and thus that the pprimary part of Kz(F) vanishes. 0

4.3.14. Remark. In fact there are no exceptional cases; Kz(F) vanishes for any finite field. To prove this for F = lF2, one can merely note that Hz(SL(3, F), z) is a finite abelian 2-group (see Exercise 4.1.28(5)), and then use Theorem 4.3.12 to deduce that Kz(F) is a 2-group. The argument in the proof of Corollary 4.3.13 then shows that Kg(F) has to vanish. With somewhat more work, Proposition 4.3.6, Proposition 4.3.11, and Theorem 4.3.12 can be turned into a proof of the following famous (and difficult) theorem of Matsumoto.

214 4. Milnor’s K2

4.3.15. Theorem (Matsumoto). IfF is any (commutative) field, Kz(F) is the free (multiplicative) abelian group Symb(F) on generators {u, w}, u, v E FX, subject to the relations of bilinearity in both variables and the relation (21, 1 - u} = 1. Proof [Hutchinson]. First of all, the given relations imply the other relations we know about, namely skew-symmetry ({u, V} = {v, u}-‘) and the relation {u, -u} = 1, since -u =

(1 - u)(l - u-1)--1,

hence

{u, -u} = (‘11, 1 - u}{u,

1 - u-1)--1 = {u-l, I- u-1) = ]

and

{u, w} = {u, w}{u, -u} = {u, -UW} = 1 uv?F1, -uw} = {uv, -uv}{w-l, -uw} = {w, -uw}-1 = {w, u}-l{v, -v}-1 = {w, u}-‘. Next, because of Corollary 4.2.18, Corollary 4.3.13, and Remark 4.3.14, the csse where F is finite is already proved. So it’s enough to show that when F is infinite, Ho(F x, Hz(GL(2, F), Z)) has the indicated presentation, and that corestriction maps this isomorphically onto Hz(GL(3, F), Z). By Lemma 4.3.10, it’s enough for the second statement to show that the corestriction map Hz(GL(2, F), Z) -+ Hz(GL(3, F), Z) is injective. We begin with the first, step, the identification of Hz(G, Z), G = GL(2, F), with the direct sum of /j2(Fx) and the group Symb(F) on symbols {u, v}, u, w E FX satisfying the indicated relations. For this we have to go back to the exact sequence (4.3.7’) in the proof of Proposition 4.3.6 and identify the image of the map H2(G, MO) + H2(B, Z) g A2(Fx x FX). We also need to use the short exact sequences (4.3.16)

0 -+ Ms --+ C’s ++ M2 + 0,

(4.3.17)

0+M4+C4%M3-$0,

and the corresponding exact sequences (4.3.16’) and (4.3.17’) in homology, in addition to (4.3.7-4.3.9). Since the orbits of G on 4-tuples and 5-tuples of distinct points in P1 (F) all have stablizer 2 g FX , one finds that H.(G,

C3)

2

@

fL(FX,

z#O, 1, m

Z>.

ix),

3. Milnor’s KZ 215 fL(G, (74) 2

@

fL(FX, Z) . (21, x2),

~l#~z,q#O,b~

and computing (dz), and (da)* as in the proof of Proposition 4.3.6 yields that (&)a : H.(G, G) --+ fL(G, C2) = fJ.(z, z) is the O-map and that (d3)+ :H.(G, C4) --+ H.(G, C3) : *. {21, z2)

Since Ho(G, Cd) + Ho(G, Ma) is surjective, the cokernel Ho(G, Mz) of the map Ho(G, M3) + Ho(G, C3) is the same as that of the map (da)* : H,(G, C4) + Ho(G, C3), i.e., Ho(G, A&) is the free abelian group P(F) on generators {x}, 2 E F \ (0, l}, subject to the relations that for ~1 # 22,

Furthermore, since (dr), coincides with the corestriction map H.(Z, Z) --+ H.(T, Z), which is a split injection, a simple diagram chase yields split short exact sequences 0 +

FX + &(G, MI ) -+ Ho(G,

0 -+ jl(F”) + H2(G, W) ---f

M2) = I’( F) --+ 0,

Hr(G, M2) + 0.

We also know that the map (do)* : H2(T, Z) --) H2(B, Z)

7 H2(T, Z)

is given by coreso(1 - w*), so when we identify H2(T, Z) with A”(T), the cokernel of (do )* can be computed to be AS2(FX) CB A2(FX). (Here AS2(FX) denotes the second antisymmetric tensor power, i.e., (FX 8

FX)I(~@~+~@4.) Now consider the commutative diagram with exact rows and columns

ffl(G, M2)

H2(B, Z)

ffo(G, M2)

Chasing the diagram, we see that the cokernel of the map H2(G, Me) + H2(B, Z), which is H2(G, Z), is the direct sum of /j2(Fx), corresponding

216 4. Milnor ’s Kz

to the split copy of FX in G, and the cokernel of a certain map P(F) -+ AS2 (F x ). Disentangling the various identifications made (see [Hutchinson, pp. 188-1901) sh ows that this map sends {z} E P(F) to (1 - z-‘) A 2-l. (A denotes the antisymmetric tensor product.) Thus Ho(FX, H2(SL(2, F), Z)) = AS2(FX x FX)/((l - z)

A

z : z E FX \ {l}),

which is exactly the group with generators (2, y} subject to bilinearity, antisymmetry, and the relation (1 - z, z} = 1. To finish the proof, it’s enough to show that the corestriction map H2(G-W, FL z) -+ Hs(GL(3, F), Z) is an injection. The proof is quite similar to that of the injectivity part of Theorem 4.3.12. As in that proof we let G = GL(3, F) and let Cj be the free abelian group on distinct (j + l)tuples of points in P2(F) such that no three are colinear if j 2 2. However, in this case we have P2 = (FX )” (the diagonal matrices) and we can take Ps to be the stabilizer of ([er], [es], [es], [er + e2 + es]), which is just the group 2 E FX of scalar matrices. The map (ds)+ : H,(P3, Z) -+ H.(P2, Z) turns out to be the O-map since Ps is central. So the proof proceeds as before, except that this time it turns out that the map Hl(G, M2) + Hl(P2, Z) is the O-map, Hl(P2, Z) 5 Hl(G, MI) S (FX)3, and the map Hz(G, MO) + Hl(G, MI) has image ” FX. Write Hz(P, Z) E Hz(GL(2, F) x FX, Z) as Symb(F) @ i(F”) @(FX @a FX) @ i(FX) , \ \ / &(FX,Z) Hz(GW, FL Z) and Hs(Pr, Z) Z Hz((F”)~, Z) as /j2((Fx)3). Then coreso(w - 1) sends (a, b, c) A (a’, b’, c’) H ({a, Klc’} - {a’, b-lc}, c A c’ - b A b’, a’~~-lc-a~~b’-lc’+c~b’+b’~c . ) Thus the cokernel of do* : Hs(Pr , Z) ---) Hs(P, Z) is isomorphic to -c’@b-b@c’,bfib’-c/id

n

Symb(F) @ A(FX). Going back to the commutative diagram with exact rows and columns H2(P1, z> 1

b*

H2(G, MO) - H2(P, z) - H2(G, Z) --% Hl(G, MO)

3. Milnor’s K2

217

we see that Hs(G, Z) is the cokernel of a certain map FX + Symb(F) $ A” (F x ). A messy diagram chase shows that this map is actually the O-map (in other words, the image of Hs(G, A&) in H2(P, Z) is contained in the image of de,), so Hs(G, Z) s Symb(F) @ /j2(Fx), as asserted. 0 4.3.18. Exercise. Show that Ks(Ri x Rs) 2 Ks(Ri) @ K2(&) for any two rings RI and RZ. 4.3.19. Exercise. This exercise concerns Ks(Z/(m)) when m is a positive

integer. (1) Show from Theorem 4.3.1 and Exercise 2.5.17 that a proof that K&z) 2 z/2 ( see Exercise 4.3.20 below) would imply that Ks(Z/(m)) has order at most 2 for any positive integer m, and would have to be generated by the Steinberg symbol { -1, -1). (2) If P is an odd prime and r > 1, R = Z/(p’) is a local ring, and the quotient of this ring by its maximal ideal is the field F, = Z/(P). Observe that RX is a group of order Pr-‘(P - l), and that the quotient map R ++ IF* induces a map RX -+ lFc which must be an isomorphism after inverting elements of order a power of P. Thus this map splits. Show also that RX contains an element of order P +‘, hence that its Sylow psubgroup is cyclic. Since lFG is cyclic of order prime to p, deduce that RX % IF; @ Z/(p’-‘) is cyclic. (3) Show by an analogue of the argument in the proof of Corollary 4.2.18 that all Steinberg symbols must be trivial for R = Z/(p’) , p an odd prime. (4) Deduce from (3)) from the Chinese Remainder Theorem, and from Exercise 4.3.18 that all Steinberg symbols are trivial for Z/(m), m odd. Deduce from (1) that a proof that Ks(Z) 2 Z/2 would imply that Ks(Z/(m))1s’ t rivial for m odd. (It is known that { -1, -1) is non-trivial in Ks(Z/(T)), r > 1.) 4.3.20. Exercise. This exercise concerns Kz(Z). A proof that Kz(Z) g Z/2 is given in [Milnor, $101. We outline here another method of attack. (1) Apply the same method of proof used in the proof of Theorem 4.3.12 to show that for any n 2 4, the corestriction map H2(SL(n, -a q -+ &(SL(n +

1, Z), Z)

is surjective. Use the action of SL(n + 1, Z) on

x=

{a E zn+l : l&21+. . . + Za,+1 = “} /{fl}.

Identify points of X with vectors in Zn+’ (up to a sign), and let Ck be the free abelian group on ordered (Ic + 1)-tuples of distinct points in X, with the extra condition that any subset consisting of < n + 1 such vectors should be a set of rows in a matrix in GL(n + 1, Z). The rest of the proof should be extremely similar to that of Theorem 4.3.12.

218

4.

(2)

Milnor’s Kz

The same ideas apply to the cases n = 2 and n = 3; however, things are more complicated because of the fact that SL(2, Z) is not perfect. In fact, it is a classical fact that SL(2, Z) is generated by the elements

s=(_4 g1 T=(J1 ;) (this follows immediately from Theorem 2.3.2 and the relations eis(l) = ST-l, ezl(l) = S-lT); furthermore, this gives a presentation of SL(2, Z) as an amalgamated free product (S, T ( S4 = T6 = 1; S2 = T3). (The freeness is proved using the action of SL(2, Z) on the upperhalf plane-see [SerreTrees, p. 351.) Thus Hi (SL(2, Z), Z) is the free abelian group on S and T satisfying the same relations, and -1 0 so is cyclic of order 12. Examining the action of ( 0 1 > On SL(2, Z), show that Hi(GL(2, Z), Z) is isomorphic to (Z/2)2, with one of generators coming from SL(2, Z). (3) Plugging the results of (2) into the argument of (l), show that H2(SJ5(7& a z> is a finite 2-group for n = 3 or 4. (In fact, it is (Z/2)2 in both cases, but the corestriction map WSL(3, z), z) -+ &(SL(4, z), z) is not an isomorphism [vandenKallen] .) (4) Deduce from (1) and (3) that KS(Z) is, up to at worst a finite 2-group, generated by the Steinberg symbol { -1, -1). Careful analysis shows in fact that there is nothing else. Since we know that this Steinberg symbol is an element of order 2 (Example 4.2.13), Ks (Z) is of order 2.

4. Applications of IT2 In this section we discuss applications of Ks in several quite different fields. First are the rather direct applications to Ki calculations that follow from the long exact sequence of Theorem 4.3.1. Then we briefly introduce the applications of Ks to number theory, which have attracted considerable recent attention. Finally, we mention some applications of Ks in analysis and topology. Computing Certain Relative K1 Groups. One of the first applications of Ks follows from Corollary 4.3.13 and Remark 4.3.14. Namely, we obtain a new proof of the following.

4. Applications of Kz

219

4.4.1. Theorem. Let R be the ring of integers in a number field, and let p be a non-zero prime ideal in R. Then SK1 (R, p) = 1. Proof. We use the exact sequence of Theorem 4.3.1: Kz(RIp) + =I(% P) + SKI(R) + SKl(R/p). Since R/p is a finite field (see the proof of Theorem 1.4.18), SK1 (R/p) vanishes by Proposition 2.2.2 and Kz(R/p) vanishes by Corollary 4.3.13 and Remark 4.3.14. So SK1 (R, p) ” SK1 (R). This vanishes by [Milnor, Corollary 16.31. While this is a hard result, vanishing of SK1 (R) is elementary when R is a Euclidean ring (Theorem 2.3.2), so for instance we obtain relatively elementary proofs of the vanishing of SK1 (R, p) when R = Z or R = Z[i] or R = Z[w]. Proving this directly is not so easy even when R = Z (the proof sketched in Exercise 2.5.17 uses Dirichlet’s Theorem on primes in arithmetic progressions). 0 Similarly, we already know from Lemma 4.3.2 that when F is a field, there is a close relationship between KZ (F) and SKI (F [t], (t2 - t)). In fact, granted the non-trivial fact (which we haven’t proved) that K2 (R) Z K2 (F) for R = F[t], Matsumoto’s Theorem for F is basically equivalent to a proof that there are no non-trivial relations (i.e., relations not consequences of the relations in Theorem 2.5.12), among the relative Mennicke symbols for SKl(F[tl, (t2 - t)). 4.4.2. Proposition [Keune]. The map d : K2(F) + SKI(R) of Lemma 4.3.2, where R = F[t] and I = (t2 - t), maps

{a, b}

I-+

[1+ (a - 1) @$ (t2 - t)

1

(1 + (b - 1)t) (t2 - t) I.

Proof. Using the notation of Lemma 4.2.15, let a(t)

= w12(a)212((a - 1)t)wn(-a),

/3(t) = w21(-u-1)221((1 - a-l>qw21(a-‘)

Then a(O) = w~~(~)wIz(--~) = 1 and P(0) = y(t) so

in St(R).

2021(-a-~)w2l(a-~) = 1. Let

= cY(t)z21(u-1)P(t)~21(-~-1)~12((~

- l)%

that y(O) = 221(~-~)221(-u-‘) = 1. Then y(l) =

w12(u)212(a - qw12(-4~21(a-1)

w2l(-u-1)221(1 =

u-1)w21(u-1)z21(-~-1)~12(~

- 1)

w12(+12(-1)~21(~-‘)212(-4~12(4

Z21(-u-l)z21(1-

u-1)~21(u-1)~12(-a)~12(u - 1)

=

w12(u)~12(-l>~21(a-1)~21(-~-1)~21(1)~12(--1)

=

wl2(~)~12(-l)~21(1)212(-1)

=

w12(a)w2(-1)

= hz(a).

220 4. Milnor’ s K2

So ifS(t) = [y(t), h13(b)], S(0) = [l, h(b)] = 1 and b(1) =

[hl2(a), h13@)1

T h u s a({a, b } ) can be computed by tracing what happens when we apply the “snake” process in the proof of Theorem 4.3.1 to 6(t) E St(R). Now = {a, b}.

(PR(‘- dt))

=

-l~~u-l,,

( -u

= (

(_,‘- 1

y)

y)

(

(;

1+ (a - 1)t -(u-l - l)“(P - t)

u:l ;) (; -“$ -

Q)

(yl)t)

(a - l)2(t2 - t) E c&q2 R) 7 > * >

(PR(6(t)) =

=

(

1 + (u - 1)t -(a-’ - 1)2(t2 - t)

(a - 1)2(P - t) * > -b(u - 1)2(t2 - t) 1 + (u - 1)t ( b-‘( a-’ -*l)2(t2 -t) > 1 - (1 - b-‘) w(t’ - t)2 (1 - b)(u - 1)2 (1 + (u - 1)t) (t2 - t) * * 1.

The result then follows after simplifying.

0

4. Applications of K2

221

Keune [Keune] used this to obtain a new proof of Matsumoto’s Theorem along the following lines: (1) First prove that there are no non-trivial relations (i.e., relations not consequences of the relations in Theorem 2.5.12), among the relative Mennicke symbols for SKi(R, I). This is done in [Bass, Ch. VI, 521. (2) Then show that Kz(F) is generated by Steinberg symbols, i.e., the natural map + : Symb(F) + Kz(F) is surjective. This uses only the easier part (surjectivity) of Theorem 4.3.12. (3) Then construct a map p : SK1(R, I) -+ Symb(F) using the presentations of the two groups. (4) Check by direct calculation that &+op is the identity on generators of SKl(R, I), using Proposition 4.4.2. (5) It follows that d and $ have to be injective, proving in particular Theorem 4.3.15. We omit the details since we have not proved the hard fact that the relations of Theorem 2.5.12 give a presentation for SKI(R, I). We see also that this fact must be of difficulty comparable to that of Matsumoto’s Theorem. K2 of Fields and Number Theory. The study of K2 of fields is intimately connected with certain questions in number theory. The reader who wants to learn more about this relationship is referred to [Milnor, $11 and §14$16] and to [Srinivas, $8 (The Mercurjev-Suslin Theorem)] for a much deeper discussion, but we will try here to sketch at least a few basic ideas. To motivate everything, recall that our proof (Example 4.2.13) of the non-triviality of { -1, - 1) E Kz (R) depended on the use of the quaternions W. In addition, as related circumstantial evidence, recall that (-1, -1) is trivial in Kz(C) (Example 4.2.19), and that there is no non-trivial finitedimensional division algebra over Cc (W @n Cc g Mz (C)) . And note as well that we have shown that Kz(lF,) = 1 for any finite field F,, while it is a classical fact due to Wedderburn that there are no non-commutative finite division algebras. All these facts suggest a close relationship between Kz(F) for a field F and the existence of non-commutative finite-dimensional division algebras over F, which is measured by the Brauer grozlp Br(F), an important invariant of the arithmetic of the field. We will see that group homology makes an appearance in this subject as well. Before getting to the quaternion and division algebras, we start with something quite classical, and in fact closely related (see [SerreCourseArith, Ch. III]). As a by-product of our work, we will obtain a proof of the Law of Quadratic Reciprocity. 4.4.3. Definition. Let F be a field of characteristic # 2. The Hilbert symbol of F is the map ( , )F : FX x FX --+ {fl} defined as follows: ifa,b E FX, (a, b)p = 1 if there exist 2, y, z E F, not all zero, such that z2 = ax2 + by2, and (a, b)F = -1 otherwise. It is clear that (a, b) only depends on the images of a and b in FX/(FX)2. (Here (FX)2 is the subgroup of FX consisting of perfect squares.) Thus the Hilbert symbol is identically 1 if every element of F is a perfect square, for instance, if F is

222 4. Milnor’s K2

algebraically closed. It is also clear that if F = Iw, (a, b)~ = 1 if and only if a and b are not both negative. 4.4.4. Lemma. Let F be a field of characteristic # 2, and let a, b E FX. The Hilbert symbol (a, b)~ is 1 if and only if a lies in the image of the norm map N : F(&)X -+ FX. Proof. If b = c2 is a perfect square in F, then F(h) = F and N is the identity, so the condition is always satisfied. But in this case c2 = a. O2 + b. l2 so (a, b)F = 1. So suppose b is not a perfect square in F. Then F(G) = {x + y& : 2, y E F} a n d N (F (&) “> = {x2 - by2 : x, y E F,

not both 0).

If a = x2 - by2, then x2 = a. l2 + by2, so (a, b)F = 1. Conversely, if there exist x, y, z E F, not all zero, such that .z2 = ax2+by2, then ax2 = z2-by2. We can’t have x = 0, since then N(z+y&) = 0 and z+y& = 0 (the norm is the product of the conjugates, so it vanishes only on the O-element), so a =

N(z+y&)

=N

X2

Thus a lies in the image of the norm map N : F( &) ’ -+ Fx .

0

4.4.5. Proposition. Let F be a field of characteristic # 2, and suppose that for any quadratic extension F(d) ofF, N(F(fi)X) has indexat most 2 in FX. Then the Hilbert symbol (a, b)F, for a, b E FX, only depends on the Steinberg symbol {a, b} E Kz(F), and defines a homomorphism KS(F) -+ {*I}. Proof. Because of Matsumoto’s Theorem (4.3.15), it’s enough to show the Hilbert symbol satisfies the relations in Symb(F). Obviously the Hilbert symbol is symmetric (or anti-symmetric, since it takes values in {fl}). If a # 0, 1, then (a, l-a)F = 1 since a.12+(1-a).12 = 12. So we have only to prove bilinearity in the first variable. If (al, b)F = 1 and (az, b)F = 1, then by Lemma 4.4.4, al and a2 lie in the image of the norm map N : F(&I)~ --+ FX, hence so does their product. Similarly, if (ai, b)F = 1 and (a2, b)F = -1 or vice versa, then one of al and a2 lies in the image of the norm map but the other does not, so their product cannot lie in the image of the norm map and (ala2, b)F = -1. Finally, if (al, b)p = (a2, b)F = -1, then b cannot be a perfect square in F, and al and a2 both represent nontrivial elements of the quotient group FX /N(F( &) ’ ). However, by the hypothesis on F, this quotient group has only two elements, so alo2 is trivial in FX/N(F(&)X) and (ala2, b)F = 1. 0 The hypothesis of Proposition 4.4.5 appears very special, but is satisfied in a non-trivial case of great interest, that of a local field.

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223

4.4.6. Theorem. Let F be a local field of characteristic # 2, that is, Iw, C:, a finite extension of thep-adic numbers Q,, or the field Pq((t)) of formal Laurent power series over a finite field F, (with q not a power of 2). Then for any non-trivial quadratic extension F(G) of F, N(F(&)X) has index exactly 2 in Fx . Proof. When F = Cc, there are no non-trivial quadratic extensions. When F = Iw, there is only one, namely Cc, and N(z) = 1.~1~ for z E Cc, so N(CX) = w:, which has index 2 in Rx. Thus we may assume F is nonarchimedean. Let R be the ring of integers in F and let p be its maximal ideal. The finite field R/p is called the residue-class field. Any quadratic extension F = F(d) of F is also a non-archimedean local field with its own ring of integers R and maximal ideal ?$3. Without loss of generality we may assume b E RX and b $ p 2. Choose generators x E R of p and ?i E l? of p. The quadratic extensions are of two types: unramified, that is, those for which [R/q : R/p] = 2, and ramified, those for which [R/p : R/p] = 1 (these are the only two possibilities since it is easy to see that [R/p : R/p] 5 [F : F] = 2). Note that FX ” {T?ZL : n E Z, u E RX} a n d F(&)X S (3% : n E Z, v E RX}. Then it turns out that in the unramified case, N(RX) = RX and N(ii) = r2u for some u E RX, whereas in the ramified case, N(?i) = 7ru for some u E RX and N(RX) is of index 2 in RX. In either case, N(F(&J)~) has index 2 in FX. To prove this, we have to do a calculation. Since we’re assuming the characteristic of F is not 2, the extension F(h) is separable with Galois group G = Gal(F(&)/F) cyclic of order 2, with generator (T : & H -x6. First suppose F = JF,((t)) with q odd. Then equating coefficients of power series shows that any element b = Cz”=, biti of R with leading coefficient bo = 1 is a perfect square, so there are only two kinds of nontrivial quadratic extensions of F: F = P,z ((t)), corresponding to taking b to be a constant power series b = bo Q! (lF,“)2 (this is the unramified case), and F = F&m)), corresponding to taking b = bit (the ramified case). Since N : lF;2 -+ IFp” ’1s surjective, it is easy to compute that N(iix) = RX and N(t) = t2 in the first case, whereas in the second case, N(m) = -bit but RX/N(RX) ” “; /(F,X)2. In either case, N(FX) has index 2 in FX. It remains to deal with the case where F is non-archimedean of characteristic 0, i.e., a finite extension of the padic numbers Q, for some p. In this case we can use the fact that the power series for the exponential and logarithm functions converge in a small enough disk and give an isomorphism of groups from some small compact open subgroup U of R to a compact open subgroup err of RX. Similarly, the exponential map gives an isomorphism from &U+&U, with, say, 01 = l+&, 132 = l-h, to an open u-invariant subgroup V of RX. View FX, Rx, and V as G-modules via the action of (T. If we consider the maps N : 2 H XCT(Z), a : x H XO(X)-~, then by Exercise 4.1.25, the chain complex whose maps are alternately N and (T gives a calculation of the G-homology, where G = { 1, a}, with H2n, n > 0, being ker N/ im CY, and with Hzn+l, n > 0, being ker CY/ im N. Note

224 4. Milnor ’s K2

that kera consists of the fixed points for cr, which just gives the intersection with F. Also, in the case of the G-module FX, imo = kerN, i.e., Hzn(G, FX ) = 1, n > 0. This is the simplest case of Hilbert’s “Theorem 90”-in this case, the proof is immediate, since obviously im o G ker N, while if N(z,, + x1&) = 3; - bx: = 1, then either xc = 1 and x1 = 0, so xe + xl& = 1, or else X0 + QJT; = (bxr + (x0 - 1) &) (bxr - (x0 - 1) &)-l = cr @Xl + (xc - 1) &) , Consider the long exact homology sequences (Proposition 4.1.9) applied to the short exact sequences of G-modules (4.4.7)

l~RX~Fx~{?i”:nEZ}~!Zll,

(4.4.8)

l-V+RX+A+l,

where A is a finite abelian group since V is open in the compact group RX. Prom (4.4.7) we obtain for n large the exact sequence (4.4.7’) &(G, Z) = 1 + &_r(G, RX) --+ Hzn_I(G, FX) + H2n-1(G, Z) 2 G + H2n-2(G, RX ) -+ Iz~~~-~(G, FX) = 1 .

Since by construction G permutes 131 and 02, Shapiro’s Lemma (Corollary 4.1.12) shows that H.(G, V) 2 H.(l, U), so the higher homology vanishes. Thus from (4.4.8) we obtain for n large the exact sequences [ (4.4.8’)

fl2n(G, V) = 1 + Hzn(G, RX) --) fJzn(G, A) + H2n-1(G, V) = 1, fJ2+1(G> V) = 1 + H~~__I(G, Rx) + H2n-1(G, A)

+ &n__2(G, V) = 1. Since A is finite, Hzn(G, A) and Hz+r(G, A) are finite and (non-canonically) isomorphic (this is a consequence of the fact that for an endomorphism of a finite abelian group, the kernel and cokernel are non-canonically isomorphic). So by (4.4.8’), Hzn(G, Rx) and Hsn_r(G, RX) are finite and non-canonically isomorphic. Substituting in (4.4.7’), we see that Hsn-r(G, FX) = ker o/ im N = FX/N(FX) has the same order as G, namely 2. One can also see from (4.4.7’) that there are two cases, the unramified case where Hs+r(G, FX) + Hsn_r(G, Z) 2 G is an isomorphism and H2+r(G, RX) = ker crlaX / im NIRX = RX /N(iix) = 1,

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225

and the ramified case where H~+_I(G, FX) + H~+_I(G, Z) z G is the O-map and H2n--I(G, ii”) = keralRX/imNIRX

= RX/N(RX)

has order 2. 0 Proposition 4.4.5 and Theorem 4.4.6 can often be used to construct nontrivial homomorphisms from K2 of a field to (51). For instance, in the case of Q, we obtain the following. 4.4.9. Theorem. K2(Q) is a direct limit of finite abelian groups, and K2(Q) 82 Z/2 is an infinite direct sum of cyclic groups of order two, one for each prime number p. The Hilbert symbol ( , )Q, of the p-adic numbers, when restricted to Q, kills the summands of Kz(Q) corresponding to primes other than p, and maps the summand corresponding to p onto {fl}. The Hilbert symbol ( , ) R of the red numbers, when restricted to Q, is given by the product formula

( , )R =

n(

7

>Qp,

p prime

The product converges in the sense that for a, b E Q”, (a, b)Qp = 1 for all but finitely many values of p. Proof (partially attributed by M&or to Tate [Milnor, 5111). By Theorem 4.3.12, K2(Q)1s. g enerated by Steinberg symbols; furthermore, by the Fundamental Theorem of Arithmetic, Q” is generated by -1 (of order 2) and by the prime numbers p (linearly independent and each of infinite order). For each positive integer m, let A, be the subgroup of Kz(Q) generated by Steinberg symbols {u, V} with u, v E Z, 1~1, 1~1 5 m. Then A, is an increasing sequence of groups and Kz(Q) = QA,. Note that Al is the subgroup generated by { -1, -l}, which we know to have order exactly 2. (It can’t have order greater than 2, but it maps to an element of order 2 in K2(W) by Example 4.2.13.) Since any integer can be factored into primes, A, = A,_1 if m is not prime. Also, A2 = Al since (2, -2) = 1 by 4.2.17(a) and (2, -1) = (2, 1 - 2) = 1 by 4.2.17(b). For p an odd prime, again {p, -p} = 1 and {p, 1 - p} = 1 by Theorem 4.2.17, so that {p, p} and {p, p - 1) coincide with {p, -l}, which has order at most 2. We claim there is a surjective homomorphism IF: + A,/A,_l, given by 2 H (2, p} mod A,_1 for 2 = 1,. . . ,p- 1. This will show APIA,_-1 is finite cyclic with order at most p - 1. Indeed, if z and y are positive integers 2 p - 1 and xy = lcp + r, where the remainder r is a positive integer 5 p - 1, then

or by bilinearity,

226 4.

Milnor’s K2

Since x, y, k, r 5 p - 1, this shows {P, r){p, SY)-~ = 1

mod &I,

mod Ap-l, and so the homomorphism is well defined. It’s surjective since {p, p} and {p, -1) coincide with {p, p-l} = {p-l, p}. By Proposition 4.4.5 and Theorem 4.4.6, ( , )Q, defines a homomorphism from Ks(Q) to {fl}. Next we show that ( , )Q~ is non-trivial on A, and, for p an odd prime, also trivial on A,_l. For the case p = 2, it’s enough (by Lemma 4.4.4) to note that -1 is not a square in Qz, and also does not lie in N(U&(fl)” ). Indeed, 01 {XY, PI = {r, P)

Q,” = (2” : n E Z} x {fl} x u, where U = {u E Z,X : u E 1 mod 4) [SerreCourseArith, 51.3.21, so -1 is not a square or a sum of two squares (i.e., a norm from Qs (&ij) in Q,” , and (-1, -1)~~ = -1. Now suppose p is an odd prime. We claim (-1, -1)~~ = +l, which will show (-1, -l)Q, is trivial on Al = As. To see this, note that

Q,” =

{P" XEZ}XlF,X

xv,

where U = {u E Zp” : u E 1 mod p} (again see [SerreCourseArith, $1.3.21). One can solve the equation x2 + y2 = -1 in IF,, since either -1 is a square mod p (when p E 1 mod 4), hence is a square in U&, or else - 1 is not a square mod p, IF*(m) = lFPz, and N : IFz2 --+ IF: is surjective. In either event, it follows from Lemma 4.4.4 that (-1, -l)q, = 1. Furthermore, p is not a square in Q,, and Q,( ,/@) is a ramified quadratic extension of Qp, so that z;lAV,(Jir)x) is of order 2 by the proof of Theorem 4.4.6. Since everything in U -+ Zg is a square, there is some positive integer k with 1 5 k 5 p - 1 such that the image of k in lFg q Zc is not in N(Zp(,/$)x), and (k, p)~, = -1. Thus ( , )Q, is non-trivial on A,. O n the other hand, if k and m are positive integers relatively prime to p, we claim that (k, m)~~ = +l. Indeed, ifm or k is a square mod p, then it is also a square in U& and this is obvious, whereas otherwise Q,(fi) is an unramified quadratic extension of Qp, so that Zi /N(Zp(fi)x) = 1 and k E N&,(&E) “) . In particular, this shows ( , )Q, is trivial on A,_I. Since APIA,-1 is cyclic and we see now that the various ( , )Q*‘s are linearly independent homomorphisms to {&}, it follows by induction on p that A, is a direct sum of cyclic groups, each of even order, one for each prime p’ < p, and that we may arrange for ( , )Q;, to be trivial except on the summand corresponding to p’. Passing to the limit, we get the desired structure theorem for Kz(Q). It remains to prove the product formula for ( , )w. Since ( , )R gives a homomorphism from Kz(Q) to {fl}, it follows from the structure theorem just proved that it must be a product of ( , )Q, ‘s for various primes 1. So we just need to check that each ( , )Q, occurs in the expansion. By bilinearity

4. Applications of K2

227

and skew-symmetry of Steinberg symbols, it’s enough to check the formula on (-1, -l}, on (-1, p} for p prime, and on {Q, p} for p and Q prime. We already know (-1, p} and {p, p} coincide in Ks(Q) and that (2, 2) = 1, so we can dispense with the generators (2, 2) and { -1, p} for p prime. We know (-1, -1)~ = (-1, -1)~~ = -1 and (-1, -1)~~ = 1 when p is an odd prime, so ( , )Q~ must occur in the expansion of ( , )w. Also, for any primes p and q, we have (q, p)~ = 1. On the other hand, given any prime 1, then either p is a square in Q , in which case (q, p)~, = 1 for any q, or else Q( 4) is a quadratic extension of Q. If this extension is unramified, which is the case if the image of p is not a square in FL, in particular if 1 is odd and p # 1, then N(Zl(JSi)X) = Zl but 1 +! N(ZL(@)~). So we see t h a t (q, P)QL = 1 for 1 odd, q # 1 and P # Z, and (q, P)Q, = (P, dQ, = -1 for p # q, q odd and p not a square mod q. If 1 = 2, then p is a square in QzX exactly when p = 1 mod 8. If p E 3, 5, 7 mod 8, then p is a square mod 2 but not a square in U& ‘, so U&(&I) is a ramified extension of Q2. In this case, for q an odd prime, q E N(Qz(@)“) exactly when q = 1 or -p mod 4. The extension Qs (a) of Q 2 is also ramified, and for q an odd prime, 47 E N(Q2(JZ)“) exactly when q E fl mod 8. We still have to compute (-1, p)Qp = (p, p)Qp for p odd. This is 1 exactly when -1 is a square mod p, which happens if and only if p = 1 mod 4. Now we can check that each ( , )Q1, 1 odd, occurs in the expansion of ( , )R. For p an odd prime, (p, p)QL = 1 except perhaps for Z = 2 and Z = p. We have (p, p)Qp = -1 exactly when p E 3 mod 4, and (p, p)Q, = - 1 exactly when p z 3, 5, 7 mod 8 and p $1 or -p mod 4, i.e., when p E 3 mod 4. So since we already know ( , )Q, occurs in the expansion of ( , )a, ( , )Q, must also occur for 1 E 3 mod 4 to give the correct value on (1, 1). Similarly, for p an odd prime, (p, 2)Q, = 1 except perhaps for Z = 2 and 1 = p. We have (p, 2)Q, = -1 exactly when p = f3 mod 8, so since we already know ( , )Q, occurs in the expansion of ( , )R, ( , )Ql must also occur for 1 E 5 mod 8 to give the correct value on (1, 2). Finally, suppose p is a prime with p E 1 mod 8. We can show by induction on p that ( , )Qp must occur in the expansion of ( , )R. Suppose inductively that ( , )QI occurs in the expansion of ( , )R for all Z < p. (To start the induction, this is true for p = 17 since no smaller prime is = 1 mod 8.) Since p E 1 mod 8, p is a square in Qzx, so (q, p)Q, = 1 for any prime q. For q # p odd, (q, p)~, = 1 except perhaps for 1 = q and/or 1 = p. Also, (q, p)Q, = -1 exactly when p is not a square mod q, and (q, p)Qp = - 1 exactly when q is not a square mod p. If there is a prime q < p for which p is not a square mod q, then since we already know ( , )Q, occurs in the expansion of ( , h ( , IQ, must also occur to give the correct value on {Q, P). So we need to show there is a prime q < p for which p is not a quadratic residue mod q. The following proof of this by contradiction is due to Gauss [Gauss, Disquisitiones Arithmeticce, $1291. Namely, let m = [&?I. Since p 2 17, 2m+ 1 < p. If p is a quadratic residue for all odd primes q < p, then it is also a quadratic residue modulo all odd prime powers < p (because of